A tennis player serves a ball horizontally, giving it a speed of 24 m/s from a height of 2.5 m. The player is 12 m from the net. The top of the net is 0.90 m above the court surface. The ball clears the net and lands on the other side. Air resistance is negligible. By what distance does the ball clear the net?

When the question says clearing the net, I assume they are asking for the difference between the vertical heights of the ball and net when it crosses over. So I found the time at which the ball crosses the net using the horizontal displacement: t = 12/24 = 0.5s

Then I solved for the vertical displacement of the ball at that time and got 1.225m, when I subtract from the net height I get 0.33m, but the answer is supposed to be 0.38m. I don't know if I'm either misinterpreting the question or doing my calculations wrong, so if someone could point me in the right direction that would be great!

2 answers

the height of the ball is
h = 2.5 - 4.9t^2
h(0.5) = 1.275, or 1.28
u = 24 m/s forever so x = 24 t
when x = 12 at net t = 0.5 s
v = - g t
v = -9.81 t
h = Hi - 4.9 t^2
h = 2.5 - 4.9 (0.25) = 2.5 - 1.225 = 1.275 above ground
1.275 - 0.9 = 0.375 m
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