# Volleyball projectile motion

1. Sep 17, 2006

### jap90

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.52 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground. At what initial speed must the ball be hit so that it lands directly on the opponent's back line? and What is the maximum height reached by the ball in this case?

I'm really stumped, if anyone can give me some help it would be greatly appreciated. thanks

2. Sep 17, 2006

### Staff: Mentor

What can you tell us about the equations of motion that apply to this problem? Assuming no air resistance, what are the vertical and horizontal components of motion? You need to show us your work in order for us to help you.

3. Sep 17, 2006

### jap90

well, I know that the height the ball has to travel in order to clear the top of the net is .91 m. I'm not sure if this even matters in this problem.
the Vxo=cos55 v
Vyo=sin55 v
the total distance the ball has to travel is 18m. Where my problem lies is with the initial height difference of 1.52 m. I don't know how to compensate for that. I've tried solving for t, but i dont know how to go about since the ball drops below its starting point. I've spent hours on this problem and still no luck. Anything you could tell me to get me started on the right path?
thanks

4. Sep 18, 2006

### Staff: Mentor

Are you familiar with these equations of motion?

$$d = d_0 + v_0 t + \frac{a t^2}{2}$$

$$v = v_0 + a t$$

One way to solve the problem is to figure out how long it takes the v-ball to travel the court distance horizontally, and how long it takes it to travel vertcially up and then back down to hit the ground. Those two times have to be equal in this problem. Use the sin and cos components for the initial velocity.