Peter, a 100 kg basketball player, lands on his feet after completing a slam dun

Question

Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jumps up again to celebrate his basket. When his feet first touch the floor after the dunk, his velocity is 5 m/s downward; when his feet leave the floor 0.50 s later, as he jumps back up, his velocity is 4 m/s upward.

A) What is the impulse exerted on Peter during this 0.50 s?

B) What is the average net force exerted on Peter during this 0.50 s?

C) What is the average reaction force exerted upward by the floor on Peter during this 0.50 s?

SHOW YOUR WORK

Explanation / Answer

A) impulse = change in momentum = 100*(4-(-5)) = 900 kgm/s

B) Force = impulse/time = 900/0.5 = 1800 N

C) Reaction = 1800 N

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.