# Physics

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 = 1.76 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 just barely makes it over the net? (Assume the volleyball is hit so that its path is parallel to the side-line as seen from an observer directly above the court, and that the volleyball is a point object.)

Again, I need some of the first steps. I can probably get it after that.

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1. You will also need to neglect air fraction to get an easy answer.

Solve for the value of V that puts the ball at the height of the net after traveling 9.0 m. (Nine meters is the half court length). You want the value of V that results in the ball being 0.67 m above the elevation where it was struck, at time T.

Vsin 55*T - (g/2)T^2 = 0.67
Vcos 55*T = 9.0

Two variables, two unknowns.

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2. Now it asks for the max. height. above the court reached by the ball. Do I use the equation Vf^2 = vi^2 + 2(g)(delta y) ?

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3. Start with the height that the ball was hit, and add an amount y such that
g y = (1/2) (Vsin 55)^2
since the ball took off at 55 degrees to horizontal.

The ball should have risen to a height higher then the net, and then passed just over it coming down

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4. Ok so I have the initial speed and the max height. Now it asks "At what initial speed must the ball be hit so that it lands directly on the opponent's back line?"
Do I use a process similar to that of the first question?

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5. yes, same process

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6. Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court?

What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???

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7. Yes, 18 m would be the length to use. The height (y change) would be MINUS the elevation where the ball was struck, since that would correspond to hitting the ground

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