For women\'s volleyballthe top of the net is 2.24m above the floor and the court

Question

For women's volleyballthe top of the net is 2.24m above the floor and the court measures9.0m on each side of the net. Using a jump serve, a player strikesthe ball at a point that is 3m above the floor and a horizontal
a) what minimum maginitude must it have if the ball is to clear thenet and b) what maximummagnitude can it have if the ball is to strike the floor inside theback line on the other side of the net? For women's volleyballthe top of the net is 2.24m above the floor and the court measures9.0m on each side of the net. Using a jump serve, a player strikesthe ball at a point that is 3m above the floor and a horizontal
a) what minimum maginitude must it have if the ball is to clear thenet and b) what maximummagnitude can it have if the ball is to strike the floor inside theback line on the other side of the net?

Explanation / Answer

   H= 2.24 m    L= 18 mt    H'= 3 mt    net height to tread before ball hits net = 3 -2.24 = .76 mt        by using eqn of kinematics.......    s= ut + .5at^2 ... eqn1    => .76 = ut - .5 x 9.8 x t^2   horizontal tread = ut= 9 mt    from above equations we get time (T) =.....    u= 9/T   this is the min velocity !!    for max velocity........tread horizontal = 18 mt...    so velocity u= 18/time    

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