Question

A clown in a circus is about to be shot out of a cannon with a muzzle velocity of 15.2m/s, aimed at 52.7° above the horizontal. The net is at the same height as the mouth of the cannon. a. How long will the clown in the air? b. How far away should his fellow clowns position a net to ensure that he lands unscathed?

Answer 1

1. Time, required to reach the final point is (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$t = \dfrac{2v_0\sin\theta}{g}$$

where $v_0 = 15.2m/s$ is the initial speed, $\theta = 52.7\degree$ is the launch angle, and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \dfrac{2\cdot 15.2\cdot \sin52.7\degree}{9.81} \approx 2.47\space s$$

2. The horizontal distacne is:

$$L = \dfrac{v_0^2\sin(2\theta)}{g} = \dfrac{15.2^2\cdot \sin(2\cdot 52.7\degree)}{9.81} \approx 22.7m$$

Answer. 1) 2.47 s, 2) 22.7 m.