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 2.95 m above the floor and a horizontal distance of 8.3 m from the net.

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

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.)


I have no idea where to start on this problem. Please help?

Explanation / Answer

to go clear the net. the height of the ball at the net position must be equal to 2.24m at least we have that 2.24=2.95-gt^2/2 v*t=8.3 so we have t=0.38(s). so v=8.3/t=21.8(m/s). ------ with maximum velocity, the horizontal distance allowed is only 8.3+9=17.3(m). we have v*t=17.3 0=2.95-gt^2/2 so t=0.78 so v=22.3(m/s)

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