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 3.03 m above the floor and a horizontal distance of 7.9 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?
___________ m/s

(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.)
___________ m/s

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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