During a tennis match, a player serves the ball at 26.7 m/s, with the center of

Question

During a tennis match, a player serves the ball at 26.7 m/s, with the center of the ball leaving the racquet horizontally 2.47 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 net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. Use g=9.81 m/s2.

Explanation / Answer

(a)Let the ball take t sec to cover 12m horizontally, By :-
distance = velocity x time
=>t = 12/26.7 = 0.449 sec
Thus the vertical height (h) covered by the ball:-
By s = ut + 1/2gt^2
=>h = 0 + 1/2 x 9.8 x (0.449)^2
=>h = 0.987m
Thus the distance between the net & ball (H) = (2.47 - 0.9) - 0.987 = 0.583m
(b) Ux = ucosA* = 26.7 x cos5* = 26.6 m/s
& Uy = usinA* = 26.7x sin5* = 2.33 m/s
Let the ball take t sec to cover 12m horizontally, By :-
distance = velocity x time
=>t = 12/26.7 = 0.449sec
Thus the vertical height (h) covered by the ball:-
By s = ut + 1/2gt^2
=>h = 2.33 x (0.449) + 1/2 x 9.8 x (0.449)^2
=>h = 2.034m
Thus the distance between the net & ball (H) = (2.5 - 0.9) - 2.034 = - 0.434m

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