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 on each side of the net. Using a jump serve, a player strikes the ball at a point that is 3.05 m above the floor and a horizontal distance of 7.7 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?
___ m/s

Explanation / Answer

vertical distance=3.05-2.24 =.81 s=a*t*t/2 t=sqrt(2*.81/10) t=.402 s now a)vmin*t=7.7 v min=7.7/.402 =19.15 m/s b)now s=3.05 t=sqrt(2*.3.05/10) t=.78 vmax*.78=(7.7+9) vmax= 21.41m/

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