Q. 100 kg basketball player, lands on his feet after completing a slam dunk and
ID: 2029270 • Letter: Q
Question
Q.
100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jumps up to celebrate. When his feet first touch the floor his velocity is 5 m/s downward, and 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 the player during this 0.50 s?
I have tried the impulse
Ft=m(Vf-Vi) and determine that there was 100 N
B. What is the average net force exerted on peter during this 0.50 s? mag and direction.
Ft=m(Vf-Vi) and get -200 kg*m downward
C. What is the average reaction force exerted upward by the floor on the player at the 0.50 s?
I am completely lost here!!!
Explanation / Answer
A. The impuse is a measure of energy (J) not force (N). in Ft=m(Vf-Vi) you should have been solving for Ft, not F. Also, his initial velocity is -5 since it is downward.....Ft=100kg(4m/s - -5m/s)=100kg(9m/s)=900J
B. Here is where the time taken comes into play. You are looking for the average force over the interval, aka the enegy divided by the time: Ft=900J ==> F=900J/t ==> F=900J/.5s ==> F=1800N upward. The player is pushing downward in order to jump, so the force on the player is opposite, ie upward.
C. The reaction force exerted by the floor is simply the equal and opposite of the force that was applied to it. The player exerted 1800N downward, therefore the floor exerted 1800N upward.
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