During a tennis match, a player serves the ball at 24.6m/s, with the center of t

Question

During a tennis match, a player serves the ball at 24.6m/s, with the center of the ball leaving the racquet horizontally 2.39m above the court surface. The net is 12.0m away and 0.900m 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 5.00degrees below the horizontal. When the ball reaches the net, what now is the distances 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 answere should be a negative number.

Explanation / Answer

time to reach the net is t=12/24.6 = 0.49s x-vel = 24.6m/s initial y-vel = 0m/s final y-vel = 0 + 9.8*0.49 = 4.802m/s initial y-height = 2.39m using v^2 = u^2 + 2*g*S we get S(final y-distance change) = (4.802^2)/(2*9.8) = 1.176m so final y-height = 2.39-1.176= 1.214m so height of centre of ball above net = 1.176 - 0.9 = 0.276m now if ball leaves 5deg below horizontal initial x-vel = 24.6 cos(5) =24.51 m/s initial y-vel = 24.6 sin(5) = 2.144 m/s now time to reach net = 12/24.51 = 0.49s final y-vel using v= u +gt ... = 2.144 + 9.8*0.49s = 6.946m/s using v^2 = u^2 + 2*g*S 6.946^2 = 2.144^2 + 2*9.8*S S = 2.23m change therefore 2.39 - 2.23 = 0.16 m therefore from net (below) distance 0.9 - 0.16 = 0.74m -0.74m.

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