What Initial Speed Is Required for a Volleyball to Just Clear the Net?

[trumnation]

Homework Statement

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.)

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:
[tex]
v^2 = v_0^2 + 2 a \Delta x
[tex]

but it doesn`t seem to work. I am thinking I have to calculate the time somehow, but I really can`t figure out how to go about doing that. Can anyone help?

 

[LowlyPion]

Homework Statement

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.)

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:

$$v^2 = v_0^2 + 2 a \Delta x$$

but it doesn`t seem to work. I am thinking I have to calculate the time somehow, but I really can`t figure out how to go about doing that. Can anyone help?

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

$$Y_{net} = Y_{serve} +V_y*t -1/2*g*t^2$$

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

 

[trumnation]

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

 

[LowlyPion]

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

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

Eliminate 1 and solve for the other.

 

[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..

 

[LowlyPion]

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..

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