Answer to Question #242537 in Physics for Emi

Question

Projectile Motion

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle below the horizontal. The service line is 11:9 m from the net, which is 0.91m high. What is the angle ∅ such that the ball just crosses the net? Will the ball land in the service box, whose outline is 6:40 m from the net?

Answer 1

170 km/h are 47.22 m/s.

Find the time for the ball to 'fall' from 2.5 m to 0.91 m at initial downward velocity of $v_y=v\sin\theta:$

$$0.91=2.5-v\sin\theta ·t -\frac{gt^2}{2}.$$

On the other hand, the time depends on the horizontal component of velocity as well:

$$t=\frac R{v\cos\theta}.$$

Substitute:

$$0.91=2.5-v\sin\theta ·\frac R{v\cos\theta} -\frac{g(\frac R{v\cos\theta})^2}{2},\\\space\\ -1.59=-R\tan\theta-\frac{gR^2}{2v^2\cos^2\theta},\\\space\\ -1.59=-R\tan\theta-\frac{gR^2}{2v^2}(\tan^2\theta+1),\\\space\\ \tan^2\theta·\frac{gR^2}{2v^2}+\tan\theta·R+\bigg(\frac{gR^2}{2v^2}-1.59\bigg)=0,\\\space\\ \tan\theta=\frac{-R\pm\sqrt{R^2-4·(gR^2/(2v^2)-1.59)·gR^2/(2v^2)}}{gR^2/v^2},\\\space\\ \tan\theta_1=7.32,\\ \tan\theta_2=-31.1.\\\theta=82.2°,$$

the angle below the horizontal is $90°-\theta=7.77°.$