During a tennis match, a player serves the ball at 25.6 m/s, with the center of the ball leaving the racquet horizontally 2.52 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the ret, what now is the distance between the center of the ball and the top of the net? Under a positive number if the ball clears the net. If the ball does not clear the net, your answer should be o negative number.
We know that the acceleration due to gravity,g=9.8m/s^2
(a) time, t=d/v
where d=distance and v=velocity
t= 12/25.6= 0.468 s
h=h0+v0t+1/2gt^2 = 2.52+0-4.9*0.468^2= 1.4467m
1.677-0.9= 0.7773= 0.54 m (Above the net top)
(b) angle, a=5 , v0=25.6*sin(5)
h=h0+v0t+1/2gt^2 = 2.52-25.6sin(5)*0.468-4.9*0.468^2= 0.40m
So the height is ,h=0.40-0.9= -0.49 m (Below net top)
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