Objects launched are projectilesObjects launched are projectiles balls, bullets, arrows, space ships…balls, bullets, arrows, space ships…
The The PATHPATH a projectile follows is the a projectile follows is the TRAJECTORYTRAJECTORY
The trajectory for a projectile is a The trajectory for a projectile is a PARABOLAPARABOLA
We will We will ignore ignore wind wind
resistancresistancee
Motion for a Projectile is Motion for a Projectile is described in terms of…described in terms of…
PositionPosition VelocityVelocity
AccelerationAcceleration All three of these are All three of these are vector quantitiesvector quantities
and will be represented with arrows like and will be represented with arrows like the previous topicthe previous topic
We must remember that We must remember that horizontalhorizontal and and verticalvertical velocities (components) of a velocities (components) of a projectile are projectile are INDEPENDENTINDEPENDENT! !
Drop a ballDrop a ball
AcceleratAccelerates es
DownwarDownwardd
Roll a BallRoll a Ball
Rolls across Rolls across
Constant Velocity (No Constant Velocity (No friction)friction)
Something thrown Something thrown through the air will have through the air will have both a down both a down (vertical)(vertical) and and across across (horizontal)(horizontal) velocity velocity
1.1. accelerate down accelerate down (due to gravity)(due to gravity)
2.2. go across at constant go across at constant velocityvelocity(we assume no air resistance)(we assume no air resistance)
Each velocity will Each velocity will NOTNOT affect the affect the
other other (Independence)(Independence) VVh h
(V(Vxx))
VVvv (V(Vyy))
VVxx is is ConstantConstant
VVyy is is changing changing
(accelerating b/c of (accelerating b/c of gravity)gravity)
What is the time of flight?What is the Vx?
EquationsEquations
PastPast
D = VtD = Vt
D= VD= Viit + ½ att + ½ at22
VVff= V= Vii + at + at
PresentPresent
X = VX = Vxxtt
Y = VY = Vyiyi t+ ½ gt t+ ½ gt22
Y = ½ gtY = ½ gt22
tt22 = = 2y2y//gg
VVyfyf = V = Vyiyi + gt + gt
Sample ProblemSample Problem A stone is thrown horizontally A stone is thrown horizontally
at +15 m/s from the top of a at +15 m/s from the top of a cliff 44m high.cliff 44m high.
1.1. How long does the stone take to How long does the stone take to reach the bottom of the cliff?reach the bottom of the cliff?
2.2. How far from the base of the cliff How far from the base of the cliff does the stone strike the grounddoes the stone strike the ground
3.3. Sketch the trajectory of the stoneSketch the trajectory of the stone
44m44m
15 15 m/sm/s
GivenGiven
VVxx = 15m/s = 15m/s
VVyiyi = 0 m/s = 0 m/s
G= G= -9.8m/s-9.8m/s22
y = -44my = -44mFormulaFormula
X = VX = Vxxtt
Y = VY = Vyyt + ½ t + ½ gtgt22
SOLUTIONSOLUTIONa. ta. t22 = 2 = 2 yy//gg
tt22 = = 2(-44)2(-44)/ / -9.8-9.8
t = t = √√9 t= 3sec9 t= 3sec
b. X= Vb. X= Vxxtt
= (15 = (15 mm//ss) (3s)) (3s)= 45m= 45m
4444mm
t= 3sect= 3sec
45m45m
Practice ProblemsPractice Problems
1.1. A stone is thrown horizontally at a A stone is thrown horizontally at a speed of speed of +5.0+5.0mm//ss from the top of a cliff from the top of a cliff 78.4m high78.4m high..
a.a. How longHow long does it take the stone to does it take the stone to reach the bottom of the cliff?reach the bottom of the cliff?
b.b. How farHow far from the base of the cliff does from the base of the cliff does the stone strike the ground?the stone strike the ground?
2.2. A steel ball rolls with a constant velocity A steel ball rolls with a constant velocity across a across a tabletop .950m hightabletop .950m high. It rolls off . It rolls off and and hits the ground +.352mhits the ground +.352m horizontally horizontally from the edge of the table. from the edge of the table. How fast was How fast was the ball rollingthe ball rolling??
Homework P. 102
1-3
Page 118 K and L
Objects Launched at Objects Launched at AnglesAngles
When we analyze a When we analyze a trajectory, we should see trajectory, we should see what happens to vertical what happens to vertical
and horizontal and horizontal components to components to
understand the motionunderstand the motion
Page 118 J
Find component velocities on the way up.
Find Resultant velocities on the way down.
Shallow AngleShallow Angle
Steep Angle
Use Vector Knowledge to Use Vector Knowledge to Solve for Solve for
VVxx & V & Vyy
The launching The launching angleangle and the and the velocityvelocity determine determine how farhow far the the
object will travel.object will travel. When the launching speed or When the launching speed or
force is the same, the angle alone force is the same, the angle alone will determine the range.will determine the range.
RANGERANGE: how far (horizontally) a : how far (horizontally) a projectile goesprojectile goes
NOTENOTE
For working with angled For working with angled trajectories, the launching height is trajectories, the launching height is
usually the same as the landing usually the same as the landing height When it is height When it is Y = 0!Y = 0!
Finding Time for angled launches
Time = 2Vy
g
Only works when landing/launching heights are the same. I call this the “hangtime” formula.
Example Problem for a projectile shot at Example Problem for a projectile shot at an angle…an angle…
A small metal ball is shot with a A small metal ball is shot with a velocity of 4.47velocity of 4.47mm//ss at an angle of at an angle of 6666°° above the horizontal. above the horizontal.
a.a. How long does it take the ball How long does it take the ball to land?to land?
b.b. How high did the ball fly?How high did the ball fly?
c.c. What was the range?What was the range?
GivenGiven
VVii = = 4.474.47mm//ss
0 = 66°0 = 66°
VVxx
VVyy4.44.477 6666
FormulaFormula
y = Vy = Vyyt + ½ t + ½ gtgt22
X = VX = Vxxtt
SolutionSolution
A.A. when land when land y=0y=0
0 = V0 = Vyyt + ½ gtt + ½ gt22
-V-Vyyt = ½ gtt = ½ gt22
t = -2Vt = -2Vyy/ g/ g
t = t = -2(4.08)-2(4.08)//-9.8-9.8
VVyy = 4.47(sin 66) = = 4.47(sin 66) = 4.084.08mm//ss
VVxx = 4.47(cos 66) = = 4.47(cos 66) = 1.82 1.82 mm//ss
= .83 sec= .83 sec
BB. max height =½t. max height =½t
y = Vy = Vyyt + ½ gtt + ½ gt22
y = (4.08)(.417) + ½ (-9.8)y = (4.08)(.417) + ½ (-9.8)(.417)(.417)22
y = 1.7 – 0.852y = 1.7 – 0.852y = 0.85my = 0.85m
CC. . Flight time Flight time = .833sec= .833sec
X = VX = Vxxt t
= 1.82m/s = 1.82m/s (.833sec)(.833sec)
X = 1.52mX = 1.52m
Practice problem 5
A softball is thrown with an initial velocity of 27m/s at an angle of 30o from the horizontal.
A. Find the total time the ball is in the air.
B. find the horizontal displacement of the ball.
C. Find the maximum height for the ball.
GivenGiven
VVxx = = 23.3823.38
VVyy = 13.5 = 13.5
g = -g = -9.8m/s9.8m/s
½ t= 1.38½ t= 1.38
Vyy3030°°
VVxx
27 m/s27 m/s
A. A. t = -2Vt = -2Vyy/ / gg
= -2(13.5)/- = -2(13.5)/- 9.89.8
= 2.76s= 2.76s
B.B. X = V X = Vxxtt
= = 23.38(2.76)23.38(2.76)
=65.5m=65.5mC.C. y = V y = Vyyt + ½ gtt + ½ gt22
= 13.5(1.38) + ½ (-9.8)= 13.5(1.38) + ½ (-9.8)(1.38)(1.38)22
18.63 + (-9.33)18.63 + (-9.33)
= 9.3m = 9.3m
Problem 6
All the given information of this problem is the same as #5. The only difference is the launch angle is now 60o.
A. find time B. Find range (X) C. Find max height (Y).
Vyy
VVxx
27 m/s27 m/s
6060°°
GivenGiven
VVxx = 13.5 = 13.5
VVyy = = 23.3823.38
g = -9.8g = -9.8
½ t = ½ t = 2.39s2.39s
A. A. t = -2Vt = -2Vyy/ g/ g
-2(23.38)/-g-2(23.38)/-g
= 4.78s= 4.78s
B.B. X = V X = Vxxt t
13.5(4.78)13.5(4.78)
=64.5m=64.5m
C.C. Y = V Y = Vyyt+ ½ gtt+ ½ gt22
= 23.38 (2.39) + ½ (-= 23.38 (2.39) + ½ (-9.8) 2.399.8) 2.3922
=27.9m=27.9m
The Range FormulaThe Range Formula
A fast way to find range when A fast way to find range when initial velocity and angle are initial velocity and angle are
given (angle should be with the given (angle should be with the horizontal)horizontal)
RR=VVii22 sin(2 0 ) sin(2 0 )
gg
Vi2 g Rg R
sin (2 0 )sin (2 0 )=Can Can also also
Find VFind Vii
ExampleExample Find the Find the rangerange for an object shot at for an object shot at
5050°° with an with an initial velocity of 5m/sinitial velocity of 5m/s
GivenGiven
VVii = 5m/s = 5m/s
0 = 50°0 = 50°
R = ?R = ?
g = 9.8g = 9.8
FormulaFormula
R = VR = Vii22 Sin 2 Sin 2
0 0
gg
SolutionSolution
= (5)= (5)22 sin sin (2x50(2x50°°) / 9.8) / 9.8
+ 24.62/ 9.8 = + 24.62/ 9.8 =
2.51m2.51m
p.104 3E #’s 5,2,1
p.114-115 30, 32, 34, 36, 39
Complimentary Angles = same ranges
450 gives the longest range for a projectile shot without air resistance. While the ranges may be the same, the time in the air won’t.
What are satellites?
Natural Artificial
How do satellites stay in orbit even though they have no
propulsion?
Isaac Newton saw the answer to this by thinking about the Moon. He also knew over 400 years ago that it would be possible to orbit the Earth if mankind could go fast enough. He also predicted mankind would travel to the Moon.
For low Earth orbit(200 to 500miles high)
Tangential Velocity of 8000m/s (17,500mph)
Vt will change with altitude.
The object (satellite) will stay in orbit. Every second it falls down 4.9m and
moves downrange (tangentially) 8000m. The resultant path mirrors the curvature of the Earth.
Vectors and Projectiles Review
Mr. Crabtree is standing on a perfectly good bridge in Zimbabwe when he decides to jump off. He is tied by his ankles to a bungee cord for safety. For 5.2 seconds, your teacher is in free fall. He jumped horizontally at 4 m/s as he left the bridge. The bridge is 150m above the Zambezi River. What was his vertical displacement at the end of the free fall? [neglect wind resistance]
Find horizontal displacement from the base of the building.
Find the time the ball is in the air
10m
20m