For women\'s volleyball the top of the net is 2.24 m above the floor and the cou

Question

For women's volleyball the top of the net is 2.24 m above the floor and the court measures 9.0 m by 9.0 m on each side of the net. Using a jump serve, a player strikes the ball at a point that is 2.99 m above the floor and a horizontal distance of 8.0 m from the net.

(a) If the initial velocity of the ball is horizontal, what minimum magnitude must it have if the ball is to clear the net?

(b) What maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net? (Assume a horizontal initial velocity, and that the ball is aimed straight at the net rather than diagonally across the court.)

Explanation / Answer

The height of the net is h = 2.24 m The height of release of the ball is y = 2.99 m The horizontal distance between point of throw and the net is x = 8 m Let v the initial horizontal speed given to the ball, The equation of motion for the ball is In Y direction: y-h = 0.5*gt^2 (For the ball to clear the net) 2.99-2.24 = 0.5*9.8*t^2 t = 0.39 s In X diresction: x = vt 8 = v*0.39 v = 20.51 m/s This is the minimum magnitude must it have if the ball is to clear the net The horizontal distance of the point of throw to the back line is x' = 8 + 9 = 17 m The equation of motion for the ball is In Y direction: y = 0.5*gt^2 (For the ball to clear the net) 2.99 = 0.5*9.8*t^2 t = 0.61 s In X diresction: x' = vt 17 = v*0.619 v = 27.87 m/s This is the maximum magnitude can it have if the ball is to strike the floor inside the back line on the other side of the net

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