During a tennis match, a player serves the ball at 24.1 m/s, with the center of

Question

During a tennis match, a player serves the ball at 24.1 m/s, with the center of the ball leaving the racquet horizontally 2.41 m above the court surface. The net is 12 m away and 0.90 m high.

a.) When the ball reaches the net, does the ball clear it?
b.) What is the distance between the center of the ball and the top of the net?
c.) Suppose that, instead, the ball is served as before but now it leaves the racquet at 4.00° below the horizontal. When the ball reaches the net, does the ball clear it?
(d) What now is the distance between the center of the ball and the top of the net?

Explanation / Answer

Hey, Insides. The question seems to leave out air resistance, but is all about gravitational pull. Gravity is the point of the question. It takes the ball about half-a-second to reach the net. How far will the ball drop in half-a-second under the influence of gravity? In part a), the ball drops g/8 meters in half-a-second, a little less than 1.25 meters. The exact drop is found from (12/24.1)^2 * g / 2 In b), the ball takes longer to reach the net, for the horizontal component of its velocity is now 24.1 * cos(5). In addition, the downward component of its initial velocity is no longer zero as in part a), but is now 24.1 * sin(5). The drop of the ball consists now of the influence of gravity and also the ball's own initial downward velocity: (12/(24.1 * cos(5)))^2 * g / 2

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