CP Projectile Test Answers

• 1. Which of the following best represents a graph of horizontal position as a function of time for balls launched horizontally in the positive x direction?

x

t

Constant forward velocity

CP Projectile Test Answers

• 1. Which of the following best represents a graph of horizontal position as a function of time for balls launched horizontally in the positive x direction?

x

t

Constant forward velocity

CP Projectile Test Answers

• 1. Which of the following best represents a graph of horizontal position as a function of time for balls launched horizontally in the positive x direction?

x

t

Constant forward velocity

CP Projectile Test Answers

• Which of the following best represents a graph of vertical position as a function of time for balls launched horizontally?

y

t

Increasing downward velocity

CP Projectile Test Answers

• Which of the following best represents a graph of vertical position as a function of time for balls launched horizontally?

y

t

Increasing downward velocity

CP Projectile Test Answers

• Which of the following best represents a graph of vertical position as a function of time for balls launched horizontally?

y

t

Increasing downward velocity

CP Projectile Test Answers

3. Which of the following graphs best represents the horizontal velocities vs time for balls launched horizontally?

Constant Forward Velocity

0 m/s

Time

CP Projectile Test Answers

3. Which of the following graphs best represents the horizontal velocities vs time for balls launched horizontally?

Constant Forward Velocity

0 m/s

Time

CP Projectile Test Answers

3. Which of the following graphs best represents the horizontal velocities vs time for balls launched horizontally?

Constant Forward Velocity

0 m/s

Time

CP Projectile Test Answers

4. Which of the following graphs best represents the vertical velocities vs time for balls launched horizontally?

Increasing downward velocity

0 m/s

Time

CP Projectile Test Answers

4. Which of the following graphs best represents the vertical velocities vs time for balls launched horizontally?

Increasing downward velocity

0 m/s

Time

CP Projectile Test Answers

4. Which of the following graphs best represents the vertical velocities vs time for balls launched horizontally?

Increasing downward velocity

0 m/s

Time

CP Projectile Test Answers

5. Which of the following graphs best represents the horizontal acceleration vs time for balls launched horizontally?

Constant forward Velocity

0 m/s2

Time

CP Projectile Test Answers

5. Which of the following graphs best represents the horizontal acceleration vs time for balls launched horizontally?

Constant forward Velocity

0 m/s2

Time

CP Projectile Test Answers

5. Which of the following graphs best represents the horizontal acceleration vs time for balls launched horizontally?

Constant forward Velocity

0 m/s2

Time

CP Projectile Test Answers

6. Which of the following graphs best represents the vertical acceleration vs time for balls launched horizontally?

Constant downward acceleration equal to 9.8 m/s2

0 m/s2

Time

CP Projectile Test Answers

6. Which of the following graphs best represents the vertical acceleration vs time for balls launched horizontally?

Constant downward acceleration equal to 9.8 m/s2

0 m/s2

Time

CP Projectile Test Answers

6. Which of the following graphs best represents the vertical acceleration vs time for balls launched horizontally?

Constant downward acceleration equal to 9.8 m/s2

0 m/s2

Time

CP Projectile Test Answers

• 7. An Olympic athlete throws a javelin at five different angles above the horizontal, each with the same speed: 20o 30 o 50 o 600 70 o Which two throws cause the javelin to land the same distance away?

• a)      20 o and 70 o

• b)      30 o and 60 o

• c)      20 o and 50 o

• d)      50 o and 70 o

• 8. Why? Projectiles fired at complementary angles produce the same horizontal range – check

projectile lab 2 results.

CP Projectile Test Answers

• 7. An Olympic athlete throws a javelin at five different angles above the horizontal, each with the same speed: 20o 30 o 50 o 600 70 o Which two throws cause the javelin to land the same distance away?

• a)      20 o and 70 o

• b)      30 o and 60 o

• c)      20 o and 50 o

• d)      50 o and 70 o

• 8. Why? Projectiles fired at complementary angles produce the same horizontal range – check

projectile lab 2 results.

CP Projectile Test Answers

• 7. An Olympic athlete throws a javelin at five different angles above the horizontal, each with the same speed: 20o 30 o 50 o 600 70 o Which two throws cause the javelin to land the same distance away?

• a)      20 o and 70 o

• b)      30 o and 60 o

• c)      20 o and 50 o

• d)      50 o and 70 o

• 8. Why? Projectiles fired at complementary angles produce the same horizontal range – check

projectile lab 2 results.

CP Projectile Test Answers

• 9. A projectile is fired at 30 degrees above the horizon. Which statement is true of the

• y – component of its velocity• a) it is greater than the x component of its velocity• b) it is less than the x component of its velocity• c) it is equal to the x component of its velocity• d) its y component of its velocity is -9.8 m/s2

• 10. Why – Any launch angle less than 45o produces greater initial horizontal launch velocities.

• Any launch angle greater than 45o produces greater vertical launch velocities.

CP Projectile Test Answers

• 9. A projectile is fired at 30 degrees above the horizon. Which statement is true of the

• y – component of its velocity• a) it is greater than the x component of its velocity• b) it is less than the x component of its velocity• c) it is equal to the x component of its velocity• d) its y component of its velocity is -9.8 m/s2

• 10. Why – Any launch angle less than 45o produces greater initial horizontal launch velocities.

• Any launch angle greater than 45o produces greater vertical launch velocities.

CP Projectile Test Answers

• 9. A projectile is fired at 30 degrees above the horizon. Which statement is true of the

• y – component of its velocity• a) it is greater than the x component of its velocity• b) it is less than the x component of its velocity• c) it is equal to the x component of its velocity• d) its y component of its velocity is -9.8 m/s2

• 10. Why – Any launch angle less than 45o produces greater initial horizontal launch velocities.

• Any launch angle greater than 45o produces greater vertical launch velocities.

CP Projectile Test Answers

11. Which of the following will have the greatest time of flight when fired from the

• ground to the ground?• a) A projectile fired at 100 m/s at 25o degrees• b) A projectile fired at 100 m/s at 35o degrees• c) A projectile fired at 100 m/s at 45o degrees• d) A projectile fired at 100 m/s at 55o degrees12. Why? The higher the launch angle the longer the

time of flight.The higher the launch angle the longer it takes the projectile to achieve zero vertical velocity at the projectiles maximum height

CP Projectile Test Answers

11. Which of the following will have the greatest time of flight when fired from the

• ground to the ground?• a) A projectile fired at 100 m/s at 25o degrees• b) A projectile fired at 100 m/s at 35o degrees• c) A projectile fired at 100 m/s at 45o degrees• d) A projectile fired at 100 m/s at 55o degrees12. Why? The higher the launch angle the longer the

time of flight.The higher the launch angle the longer it takes the projectile to achieve zero vertical velocity at the projectiles maximum height

CP Projectile Test Answers

• 13. Which of the following will travel the greatest horizontal distance when fired from the

• ground to the ground?• a) A projectile fired at 100 m/s at 10o degrees• b) A projectile fired at 100 m/s at 20o degrees• c) A projectile fired at 100 m/s at 50o degrees• d) A projectile fired at 100 m/s at 70o degrees• e) A projectile fired at 100 m/s at 80o degrees• 14. Why? A 45 degree launch angle results in the

greatest horizontal distance. A 500 launch angle is the closest to 450 of the choices listed.

CP Projectile Test Answers

• 13. Which of the following will travel the greatest horizontal distance when fired from the

• ground to the ground?• a) A projectile fired at 100 m/s at 10o degrees• b) A projectile fired at 100 m/s at 20o degrees• c) A projectile fired at 100 m/s at 50o degrees• d) A projectile fired at 100 m/s at 70o degrees• e) A projectile fired at 100 m/s at 80o degrees• 14. Why? A 45 degree launch angle results in the

greatest horizontal distance. A 500 launch angle is the closest to 450 of the choices listed.

CP Projectile Test Answers

• 13. Which of the following will travel the greatest horizontal distance when fired from the

• ground to the ground?• a) A projectile fired at 100 m/s at 10o degrees• b) A projectile fired at 100 m/s at 20o degrees• c) A projectile fired at 100 m/s at 50o degrees• d) A projectile fired at 100 m/s at 70o degrees• e) A projectile fired at 100 m/s at 80o degrees• 14. Why? A 45 degree launch angle results in the

greatest horizontal distance. A 500 launch angle is the closest to 450 of the choices listed.

CP Projectile Test Answers

15. A projectile is fired at 45 degrees above the horizon. Which statement is true of the y – component of its acceleration

a) the y – component of its acceleration at its maximum height is 0 m/s2

b) the y – component of its acceleration initially is 9.8 m/s2 sin 450

c) the y – component of its acceleration initially is -9.8 m/s2 sin 450

d) the y – component of its acceleration at its maximum height is -9.8 m/s2

16. Why? The y – component of the acceleration due to gravity is a constant -9.8m/s2 regardless of the y position of the projectile. It is -9.8m/s2 on a projectiles way up, at itsmaximum height and on its way down.

CP Projectile Test Answers

15. A projectile is fired at 45 degrees above the horizon. Which statement is true of the y – component of its acceleration

a) the y – component of its acceleration at its maximum height is 0 m/s2

b) the y – component of its acceleration initially is 9.8 m/s2 sin 450

c) the y – component of its acceleration initially is -9.8 m/s2 sin 450

d) the y – component of its acceleration at its maximum height is -9.8 m/s2

16. Why? The y – component of the acceleration due to gravity is a constant -9.8m/s2 regardless of the y position of the projectile. It is -9.8m/s2 on a projectiles way up, at itsmaximum height and on its way down.

CP Projectile Test Answers

15. A projectile is fired at 45 degrees above the horizon. Which statement is true of the y – component of its acceleration

a) the y – component of its acceleration at its maximum height is 0 m/s2

b) the y – component of its acceleration initially is 9.8 m/s2 sin 450

c) the y – component of its acceleration initially is -9.8 m/s2 sin 450

d) the y – component of its acceleration at its maximum height is -9.8 m/s2

16. Why? The y – component of the acceleration due to gravity is a constant -9.8m/s2 regardless of the y position of the projectile. It is -9.8m/s2 on a projectiles way up, at itsmaximum height and on its way down.

CP Projectile Test Answers

• 17 . You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, its time of flight will be

a. 1.4 times as much.

b. half as much.

c. twice as much. vfy=viy + gt t=vfy-viy t related to vi

g

d) Four times as much

e. the same

CP Projectile Test Answers

• 17 . You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, its time of flight will be

a. 1.4 times as much.

b. half as much.

c. twice as much. vfy=viy + gt t=vfy-viy t related to vi

g

d) Four times as much

e. the same

CP Projectile Test Answers

• 17 . You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, its time of flight will be

a. 1.4 times as much.

b. half as much.

c. twice as much. vfy=viy + gt t=vfy-viy t related to vi

g

d) Four times as much

e. the same

CP Projectile Test Answers

• 17 . You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, its time of flight will be

a. 1.4 times as much.

b. half as much.

c. twice as much. vfy=viy + gt t=vfy-viy t related to vi

g

d) Four times as much

e. the same

CP Projectile Test Answers

• 17 . You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, its time of flight will be

a. 1.4 times as much.

b. half as much.

c. twice as much. vfy=viy + gt t=vfy-viy t related to vi

g

d) Four times as much

e. the same

CP Projectile Test Answers

• 18. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, the maximum height it will achieve is 

a. 1.4 times as much.

b. half as much.

c. twice as much.

d. four times as much The ball will be in the air for twice as long and it will travel twice as fast initially. twice the time at twice the speed.

e. the same

CP Projectile Test Answers

• 18. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, the maximum height it will achieve is 

a. 1.4 times as much.

b. half as much.

c. twice as much.

d. four times as much The ball will be in the air for twice as long and it will travel twice as fast initially. twice the time at twice the speed.

e. the same

CP Projectile Test Answers

19. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, the vertical velocity at the maximum height will be 

a. 1.4 times as much.b. half as much.c. twice as much.d. four times as muche. the same – All projectiles at their maximum height will

have a vertical velocity of 0 m/s

CP Projectile Test Answers

19. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, the vertical velocity at the maximum height will be 

a. 1.4 times as much.b. half as much.c. twice as much.d. four times as muche. the same – All projectiles at their maximum height will

have a vertical velocity of 0 m/s

CP Projectile Test Answers

19. You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity

• it had on your first throw, the vertical velocity at the maximum height will be 

a. 1.4 times as much.b. half as much.c. twice as much.d. four times as muche. the same – All projectiles at their maximum height will

have a vertical velocity of 0 m/s

CP Projectile Test Answers

• 20. Projectiles that are fired down at 10 degrees below the horizon will have

a) a higher y velocity when they hit the ground when compared to projectile fired up at 10 degrees above the horizon at the same speed

b) a lower y velocity when they hit the ground when compared to a projectile fired up at 10 degrees above the horizon at the same speed

c) the same y velocity when they hit the ground when compared to a projectile fired up at 10 degrees above the horizon at the same speed.

d) have a velocity that could be higher or lower when compared to a projectile fired up at 10 degrees above the horizon depending on the whether the initial velocities are above 10 m/s or below 10 m/s.

CP Projectile Test Answers

• 20. Projectiles that are fired down at 10 degrees below the horizon will have

a) a higher y velocity when they hit the ground when compared to projectile fired up at 10 degrees above the horizon at the same speed

b) a lower y velocity when they hit the ground when compared to a projectile fired up at 10 degrees above the horizon at the same speed

c) the same y velocity when they hit the ground when compared to a projectile fired up at 10 degrees above the horizon at the same speed.

d) have a velocity that could be higher or lower when compared to a projectile fired up at 10 degrees above the horizon depending on the whether the initial velocities are above 10 m/s or below 10 m/s.

CP Projectile Test Answers

• The projectile fired up at 10 m/s 10o above the horizon will eventually become a 10 m/s projectile fired down

Fired up becomesFired down.

Fired down

CP Projectile Test Answers

• The projectile fired up at 10 m/s 10o above the horizon will eventually become a 10 m/s projectile fired down

Fired up becomesFired down.

Fired down

CP Projectile Test Answers

• The projectile fired up at 10 m/s 10o above the horizon will eventually become a 10 m/s projectile fired down

Fired up becomesFired down.

Fired down

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

a) What is the time of flight of the dropped ball? • Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)• t = .49s

Projectile Test-a) What is the time of flight of the dropped ball?

• Given Find FRS• g= - 9.8 m/s2 t=?s yf = yi + viyt + ½ gt2 • viy=0m/s yf = yi + ½ gt2 • yi=1.20 m yf - yi = ½ gt2

• yf=0m yf - yi = t2

• ½ g

• yf - yi = t = 0m-1.20m

• ½ g ½ (-9.8m/s2)

a) t = .49s

Projectile Test-time of flight for yi=1.20m and yf=0m

b) What is the time of flight for the ball fired at 10 m/s horizontally?

The time of flight for the horizonatally fired ball is also 0.49s. The time of flight is independent of the horizontal velocity.

Projectile Test-time of flight for yi=1.20m and yf=0m

b) What is the time of flight for the ball fired at 10 m/s horizontally?

The time of flight for the horizonatally fired ball is also 0.49s. The time of flight is independent of the horizontal velocity.

Projectile Test-time of flight for yi=1.20m and yf=0m

b) What is the time of flight for the ball fired at 10 m/s horizontally?

The time of flight for the horizonatally fired ball is also 0.49s. The time of flight is independent of the horizontal velocity.

Projectile Test-time of flight for yi=1.20m and yf=0m

c) What is the horizontal distance of the ball fired at 10m/s?

c) vix=vx= x t vx = x = .49 s (10 m/s) = 4.9 m

t

Projectile Test-time of flight for yi=1.20m and yf=0m

c) What is the horizontal distance of the ball fired at 10m/s?

c) vix=vx= x t vx = x = .49 s (10 m/s) = 4.9 m

t

Projectile Test-time of flight for yi=1.20m and yf=0m

c) What is the horizontal distance of the ball fired at 10m/s?

c) vix=vx= x t vx = x = .49 s (10 m/s) = 4.9 m

t

Projectile Test-time of flight for yi=1.20m and yf=0m

c) What is the horizontal distance of the ball fired at 10m/s?

c) vix=vx= x t vx = x = .49 s (10 m/s) = 4.9 m

t

Projectile Test-time of flight for yi=1.20m and yf=0m

c) What is the horizontal distance of the ball fired at 10m/s?

c) vix=vx= x t vx = x = .49 s (10 m/s) = 4.9 m

t

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2 yf = 1.20 m + 0 m/s (.25s) + ½

(-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2 yf = 1.20 m + 0 m/s (.25s) + ½

(-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2 yf = 1.20 m + 0 m/s (.25s) + ½

(-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2 yf = 1.20 m + 0 m/s (.25s) + ½

(-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2 yf = 1.20 m + 0 m/s (.25s) + ½

(-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2

yf = 1.20 m + 0 m/s (.25s) + ½ (-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2

yf = 1.20 m + 0 m/s (.25s) + ½ (-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2

yf = 1.20 m + 0 m/s (.25s) + ½ (-9.8m/s2)(.25s)2 = .31 m

Projectile Test-time of flight for yi=1.20m and yf=0m

d) What is the horizontal and vertical positions of the ball fired 10 m/s horizontally .25 seconds into its flight?

vix=vx= x t vx = x = .25 s (10 m/s) = 2.5 m

t

yf = yi + viyt + ½ gt2

yf = 1.20 m + 0 m/s (.25s) + ½ (-9.8m/s2)(.25s)2 = .31 m

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• Givens

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• a) vox = vocos m/s cos 65o = 4.23 m/s

• b) voy = vosin m/s sin 65o = 9.06 m/s

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• c) Given

• yo= .25 m

• y=.98m

• vo = 10.0 m/s

• g = - 9.8 m/s2

• vfy=0m/s

• vfy=viy + gt t=vfy-viy = 0 m/s – 9.06 m/s =.92seconds

• g 9.8m/s2

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

• Determine the maximum height achieved by the ball.

• d) yf = yi + viyt + ½ gt2

• yf = .25m + 10 m/s (.92s) + ½ (-9.8m/s2)(.92s)2 = 5.25m

• Determine the horizontal position of the ball at the maximum height.

• e) vix=vx= x t vx = x = .92 s (10 m/s) = 9.2 m

• t

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

f) Determine the total time of flight from the .25 m vertical launch position to the .98 m vertical landing position.

• y = yo + vosin t + ½ g t2

• .98= .25m + 10m/s sin(650) t + ½ (-9.8 m/s2) (t)2

•  

• 0= -.73 m + 9.06 m/s ( t ) + ½ (-9.8 m/s2) (t)2

•  

• A = -4.9 B = 9.06 C = -.73 m

•  

• Using the quadratic equation t = .0844s (way up ) and 1.76s (way down)

Projectile Test Answers

g) Determine the horizontal landing position of the ball.

vox = x

t

 

t vox = x = 1.76s (4.23 m/s ) = 7.44m

Projectile Test Answers

g) Determine the horizontal landing position of the ball.

vix = x

t

 

t vox = x = 1.76s (4.23 m/s ) = 7.44m

Projectile Test Answers

g) Determine the horizontal landing position of the ball.

vix = x

t

 

t vox = x = 1.76s (4.23 m/s ) = 7.44m

Projectile Test Answers

g) Determine the horizontal landing position of the ball.

vix = x

t

 

t vox = x = 1.76s (4.23 m/s ) = 7.44m

Projectile Test Answers

g) Determine the horizontal landing position of the ball.

vix = x

t

 

t vox = x = 1.76s (4.23 m/s ) = 7.44m

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0

Projectile Test Answers

h) Determine the angle of the ball as it lands in the can. (extra credit)

vfy=viy+ gt

vfy= 9.06 m/s + (-9.8m/s2) (1.76s)

vfy=-8.19 m/s

vx=4.23 m/s

Inv tan ( vfy/vx) = angle with respect to the horizon

Inv tan ( -8.19 m/s / 4.23 m/s ) = 62.7 0