A 63.3-kg basketball player jumps vertically and leaves the floor with a velocit

Question

A 63.3-kg basketball player jumps vertically and leaves the floor with a velocity of 1.95 m/s upward. (Take upward as positive and forward as positive.)

(a) What impulse does the player experience?

-Magnitude___ N.s

Direction: upward, downward, forward, backward, magnitude is zero.

(b) What force does the floor exert on the player before the jump?

-magnitude N

Direction:upward, downward, forward, backward, magnitude is zero.

(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?

Magnitude N

Direction: upward, downward, forward, backward, magnitude is zero.

Explanation / Answer

a) IMpulse = m(vf-vi) = change in momentum = (68.1)( 0- 1.95 ) = 132.8 Ns

b) Force= mg = 68.1 x 9.8 = 667.4N


c) F=132.8/0.45 = 295.1 N

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