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 12.0 m/s at an angle 49.0 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?

Answer 1

You need to use right triangle trig for the components of a two dimensional vector as follows



Think about what is happening. The ball is thrown up with some velocity and continues in that direction but more slowly as the acceleration due to gravity acts. Eventually there will be a point when the velocity in the y-direction is zero which will be the maximum point. So you can do this in one of two ways. Find how long it takes for the ball's velocity to be zero and then plug that time into the kinematic equation for displacement or you can develop a function for max height. The process for the second would be as follows:
equation for velocity in the y direction for projectile motion
\[v_{fy}=v_{i} \sin \theta-gt\]
as previously stated this would be equal to 0 at max height
\[0=v_{i} \sin \theta-gt\]
\[v_{i} \sin \theta=gt\] divide by g to get an expression for time
\[t=sin\theta/g\]
Now plug this into the displacement equation in the y direction because this is the amount of time it takes for the ball to reach its maximum height
\[y=y_{i} + tv_{iy}sin\theta-(1/2)gt^2\] substitute for t
\[y=y_{i} + (sin\theta/g)v_{iy}sin\theta-(1/2)g(sin\theta/g)^2\]
if you solve this it will give you the expression for the maximum height based on the initial velocity. From all this you should follow this general problem solving method for the last part of your problem