Question

Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jump 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. What is the average reaction force exerted upward by the floor on Peter during this 0.50 s?

Answer 1

1800 N

Solution:

Impulse = mΔv = m * (u - v) .

here m = 100 kg

u = 4 m/s

v = -5 m/s

impulse = 100 x ( 4 - ( -5 ) ) = 900 Kg m/s .

Average reaction Force ( Favg ) = impulse / Δt

Average reaction Force ( Favg ) = 900kg·m/s / 0.5s

Average reaction Force ( Favg ) = 1800 N