A 57 kg basketball player jumps vertically upward to make a shot. At the very top of her jump, she throws the basketball with a speed of 4.4 m/s.

Question

What is her recoil velocity immediately after she releases the ball? The mass of the basketball is 0.62 kg. At the very top of the jump, what is the change in KE of the basketball? What force would the player exert on the ball if she extends her arms 0.5 meters to shoot the ball?

Answer 1

What is recoil velocity (vr)?

At top of jump, the player and ball are motionless for an instant so their total momentum is 0. An instant later the throw is made and the ball is moving one direction at 4.4 m/s. Its momentum (momB) is mv = (0.62 kg)(4.4 m/s) = 2.728 kgm/s. Momentum of the player (momP) in the opposite direction is (57 kg)vr. Total momentum of both after the throw must still be 0 which means momB + momP = 0; momP = 0 - momB; 57vr = -2.728;

vr = -0.048 m/s

What is change in ball's KE?

It starts with 0 KE because it has no velocity at the top of the jump. It ends at 4.4 m/s which gives it

KE = 1/2mv² = (0.5)(0.62)(4.4)² = 6.0 J

So the change in KE is 6.0 J

What force (F) is exerted by the player?

The change in KE is equal to the work done on the ball. Work = F x the distance the force was exerted over (d). d here is how far she extends her arms to throw the ball = 0.5 m.

So 6.0 J = F(0.5)

F = 12 N