# Projectile Motion - Volleyball

1. Sep 29, 2008

### trumnation

1. The problem statement, all variables and given/known data

A regulation volleyball court is L = 18.0 m long and the net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.61 m directly above the back line, and the ball's initial velocity makes an angle θ = 49° with respect to the ground. At what initial speed must the ball be hit so that it just barely makes it over the net? (Assume the volleyball is a point object and is hit so that its path is parallel to the sideline, as seen from directly above the court.)

3. The attempt at a solution

I am really lost at how to figure out this one without time. I tried using the following equation since you are given the height it has to reach, and the acceleration (-9.8) and I think the final velocity would be zero:
$$v^2 = v_0^2 + 2 a \Delta x [tex] but it doesnt seem to work. I am thinking I have to calculate the time somehow, but I really cant figure out how to go about doing that. Can anyone help? 2. Sep 29, 2008 ### LowlyPion It's not quite so simple. First you need to determine the Vertical and Horizontal components of velocity based on the 49 degrees. Then you can develop the x and y motion equations using the 18 m as the range and the height of the net as the height at the instant of clearance. The equations [tex] Y_{net} = Y_{serve} +V_y*t -1/2*g*t^2$$

$$X_{to.net} = V_x*t$$

3. Sep 29, 2008

### trumnation

Ok, how though, do I find the time? Or should I solve for t in the x equation and sub it in?

4. Sep 29, 2008

### LowlyPion

You should have equations with only initial velocity and time unknown.

Eliminate 1 and solve for the other.

5. Sep 29, 2008

### trumnation

Ok so I tried that, and I think I have an error somewhere because it doesn't work. Here is what I did:

t = 9/(Vsin49)

2.43 = 1.61 + Vcos49t - 0.5*9.8t^2
0.82 = Vcos49 * 9/Vsin49 - 4.9 * (9/Vsin49)^2

V = 7.9 m/s

I can't seem to find what I did wrong though..

6. Sep 29, 2008

### LowlyPion

I calculate your equation differently.

Using Sin49 = .755 and Cos49=.656
I get: t=11.92/v

Substituting:
.82 = .656(11.92) - 4.9(11.92/V)2
.82 = 7.82 - 4.9(11.92/V)2
7 = 4.9(11.92/V)2
V2 = 4.9(11.92)2/7 = 99.46
V = 9.97 m/s