Basketball player dunk - impulse, net force, GRF

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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
 
on Phys.org
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?


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