Problem 4-16 - Projectile - Part 13

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 and the top of the net is \(0.90 \;m\) above the court surface.

(a) How long is the ball in the air, assuming it clears the net and lands in the other court?
(b) How far horizontally does the ball travel?
(c) With what velocity does the ball strike the court surface?
(d) By how much distance does the ball clear the net?

Accumulated Solution

\(y = v_{0,y}t + (1/2)a_yt^2 \\ a_y = -9.8\; m/s^2 \\ y = -2.5\; m \\ t = 0.714\; s = 0.71\; s \; \text{(answer to (a))} \\ x = 17\; m \; \text{(answer to (b))} \\ v_x = 24\; m/s \\ v_y = -7.00\; m/s \\ v = 25\; m/s\; \text{at} \;16^\circ \; \text{below the horizontal (answer to (c))} \\ \text{time to net} = 0.5\; s \\ y = 1.23\;m\; \text{(fallen in} \; 0.5\; s)\)

Height above the ground \(= 2.5 \;m - 1.23 \;m = 1.28\; m\)

The ball clears the net by:
\(1.28 - 0.9 = 0.38 = 0.4\; m\)

You have completed this problem.