contestada

A 65.1-kg basketball player jumps vertically and leaves the floor with a velocity of 1.82 m/s upward.
(a) What impulse does the player experience?
(b) What force does the floor exert on the player before the jump?
(c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?

Respuesta :

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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