Question

Currently, your committee is investigating the high incidence of ankle injuries on the basketball team. You are watching the team practice, looking for activities which can result in large horizontal forces on the ankle. Observing the team practice jump shots gives you an idea, so you try a small calculation. A 40-kg student jumps 1.0 meters straight up and shoots the 0.80-kg basketball at his highest point. From the trajectory of the basketball, you deduce that the ball left his hand at 30° from the horizontal at 20 m/s. What is his horizontal velocity when he hits the ground?

Can a diagram be drawn? and how the formulas were derived be shown also?

Answer 1

use conservation of momentum.
Before impact, momentum of both ball and player in horizontal direction is 0
After impact momentum of the ball in horizontal direction is 0.8 * 20 * cos 30^o
Momentum of the player in the same horizontal direction is 40 * V
Hence total 40 V + 13.8564
By conservation this has to be equated to 0
Hence V = - 13.8564 / 40 = -0.3464 m/s
negative sign indicates that the player would go back after he hits the ball.

in case of any doubt, please do comment sir.