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

Answer 1

Step-by-step explanation:

Impulse = change in momentum

Initial momentum = mass x initial velocity = 100 x 5 = 500 kg m/s

final momentum = mass x final velocity = 100 x - 4 = -400 ( - ve sign due to reversal of direction )

change in momentum = final momentum - initial momentum

= - 400 - 500 = - 900 kg m/s .

As it is - ve , it acts upwards .

So magnitude of impulse on Perter = 900 kg m/s