Basketball player dunk - impulse, net force, GRF

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SUMMARY

The discussion focuses on the physics of a basketball player's landing and subsequent jump, specifically analyzing the average net force, reaction force, and impulse during a 0.50-second interval. Peter, a 100 kg player, experiences a downward velocity of 5 m/s upon landing and an upward velocity of 4 m/s when jumping. The average net force exerted on Peter is calculated as -200 N, while the average reaction force from the floor is 1181 N. The impulse, which is the change in momentum, requires further clarification and calculation based on the defined parameters.

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antonystad
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Homework Statement


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


Homework Equations



a) Fave = mass *(Vf - Vi) / time ?
b) Fgrf = mass *(Vf - Vi) / t + mg ?
c) no idea

The Attempt at a Solution



a) 100 (4-5) / 0.5 = -200kg
b) 100 (4-5) / 0.5 + 100 * 9.81 = 1181 N
c) HELP ME!

Am I on the right track? Please help! Many many thanks in advance
 
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antonystad said:
a) 100 (4-5) / 0.5 = -200kg

Your units are wrong. (kg(m/s))/s does not equal kg, it equals Newtons

Anyway you forgot to include the direction. The "4m/s" and "5m/s" have opposite directions.

One way to look at is, what is the total change in momentum?
He starts with 500(kg*m/s) downwards and ends up with 400(kg*m/s) upwards, so what is the change in momentum?
(And change in momentum is equal to average force multiplied by time, so then what is the average force?)


For part (c)
What is the definition of impulse?


P.S.
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