Answer to Question #339637 in Mechanics | Relativity for Chenay

Question

A tennis ball of mass 0.0551kg is served. It strikes the ground with a velocity of 54.0m/s at an angel of 22° below the horizontal. Just after the bounce it is moving at 53.0m/s at an angel of 18° above the horizontal. If the interaction with the ground lasts 0.0610 s, what is the magnitude of the average force exerted by the ground on the ball?

Answer 1

$F=\frac pt=\frac m{2t}(v_1\cos\alpha+v_2\cos\beta)=42.0~N.$