# Problem 4-16 - Projectile - Part 10 - (B)

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

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

Correct.

$\mathrm {\theta = \tan^{-1} |v_y|/|v_x| = \tan^{-1}\; 7/24 = 16^\circ}$

Here is a picture that includes the net. The net is 12 m from the origin horizontally. Before continuing calculate how long it takes for the ball to arrive above the net.

Continue.