(a) What is the minimum speed the ball must have to clear the net?

Question

2. In a women’s volleyball court the court measures 9.0 m by 9.0 m on each side and the top of the net is 2.24 m off the ground. A tall athlete serves the ball at 3.0 m above the ground and 8.0 m from the net. If the initial velocity of the ball is horizontal,
(a) What is the minimum speed the ball must have to clear the net?
(b) What is the maximum speed the ball can have to ensure it is not out when it hits the ground beyond the net?

Answer 1

(a)ball falls 3-2.24m = .76 m in

t = sqrt(2*.76/9.8) = 0.394 s

8 = v*0.394

So, v = 20.31m/s

(b) to drop 3 m t = sqrt(2*3/9.8) = 0.783 s

To cover 17m (8+9) v = 17/0.783 m/s = 21.73m/s

Answer 2

The ball must not fall more than s = 3.0 - 2.24 m by the time it travels 8.0m, The time it takes to travel horizontally 8.0 m is t = 8.0/v0, where v0 is the serve velocity. The ball falls at a rate of s = .5*g*t^2, where g = accel of gravity, or 9.8m/sec^2. Taking t from the first equation to get s = .5*g*(d^2/v0^2). Solve for v0. For the second part, the ball must fall 3.0 m before it reaches the foul line, which is 8.0m+9.0m from the server. The equations are the same.