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lorasvet [3.4K]
3 years ago
10

A 65.1-kg basketball player jumps vertically and leaves the floor with a velocity of 1.82 m/s upward.

Physics
1 answer:
loris [4]3 years ago
3 0
Impulse = change in momentum
J = mΔv = (65.1 kg)(1.82 m/s) = 118.5 kg-m/s

the floor exerts the normal force, which is equal to the player's weight
N = W = mg = (65.1 kg)(9.8 m/s²)

Impulse = average force * time applied
118.5 kg-m/s = F (0.450 s)
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