Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 24 m/s at an angle 35 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

g = 9.81m / s^2

(1) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 23 m/s when it reaches a maximum height of 13 m above the ground.

What is the speed of the ball when it leaves Sarah's hand?

How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

(2) After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 13 m above the ground, and takes 2.399 s to get directly over Julie's head.

What is the speed of the ball when it leaves Sarah's hand?

How high above the ground will the ball be when it gets to Julie?

Answer 1

1) Julie throws a ball from an initial height h₀ = 1.5 m with a speed of v = 24 m/s at an angle of θ=35^∘ above the horizontal. Sarah catches the ball when it is again at a height h = 1.5 m.

the speed of the ball when it leaves Sarah's hand is

The maximum height is achieved when the vertical speed is zero. In the vertical direction, the ball has an acceleration of

a= g =−9.8 m/s2, which is negative because it is downward. The vertical speed vy, as a function of height h, is given by

where, in our case, v0=vsin(θ)=24sin(35∘)=13.77 m/s. When vy=0, we can solve for h to find

This is actually the change in height of the ball, so we need to add back in the initial height. That means the maximum height is

  

  

2)

  

Then for the vertical component

So the speed is,

The height at that time is given by