Motion in 2-Dimensions

Projectile Motion

A projectile is given an initial force and is then (assuming no air resistance) is only acted on by gravity.

The path it takes is called the trajectory.

Path is a parabolic shape.

Horizontal ProjectilesHorizontal - cause is

due to the throw or movement.

vh is constant. The force only acts for an instant. No force = no acceleration.

dh = vht

Vertical - force acting on the object is only gravity.

vv changes because gravity is acting on the object.

dv = vit + 1/2gt2

Time is interchangeable between horizontal and vertical directions

Motion Map

Practice

A ball is rolled horizontally off of a flat roof. It hits the ground 2.3m away from the wall. If the ball was in the air for 1.2 seconds, what was the horizontal velocity of the ball just before it rolled off of the roof?

How tall was the roof?

Projectiles Launched at an Angle

o When launched at an angle the projectile has an initial vertical component as well as a horizontal component.

o 1st step should be to find the components of the velocity vector.

o These act independently from each othero Solving for maximum height use half of the trajectory o vertical velocity = 0 and t = 1/2 total to Range - how far the object travels horizontally.o Flight Time (hang time) - the total time the object is in

the air.

Motion Map

VerticalVelocity

Horizontal Velocity

Vertical Accelerationand Fg

Maximum Range

• A projectile will have the maximum range when it is fired at an angle of 450

• Two angles that are the same value away from 45 will have the same range. • EX 30 and 60, 10 and 80

• Animation 1

Practice

• A ball is launched with an initial velocity of 4.47m/s at an angle of 66o above the horizontal.

• a) What is the maximum height the ball attained?

vvi=visin vvi=4.47m/s sin(66) = 4.08m/s

vh=vicos vh=4.47m/s cos(66) = 1.81m/s

We need to know how long the ball takes to reach its maximum height.

vvf=vvi + at

0 = 4.08m/s + (-9.8m/s2)t

t = .416s

Now we know how long it takes to reach maximum height.

dv = vvit + 1/2gt2

dv = (4.08m/s)(.416s) + 1/2(-9.8m/s2)(.416s)2

dv = .849m

How long does it take to return to the height it was launched from?(total time)

Time to max height = 0.416s

Total time = 2 x 0.416s

Total time = 0.832s

What is the range?

• Range(R) = dh

• dh = vht

• R = (1.81m/s) (.832s)• R = 1.51m

#2

A soccer player kicks a ball into the air at an angle of 36 degrees. The initial velocity is 30m/s. How long is the ball in the air? What is the horizontal distance traveled by

the ball? What is the maximum height reached by

the ball?

#3

An arrow is shot at 30 degrees with a velocity of 49m/s and hits a target. What is the maximum height attained by

the arrow? The target is at the height the arrow was

shot from. How far away is the target?