A force platform is a tool used to analyze the performance of athletes measuring

Question

A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on the ground as a function of time. Starting from rest, a 59.0 kg athlete jumps down onto the platform from a height of 0.720 m. While she is in contact with the platform during the time interval 0 < t < 0.8 s, the force she exerts on it is described by the function below. F = (9 200 N/s)t - (11 500 N/s2)t2 Assume the positive y-axis points upward. (a) What was the athlete's velocity when she reached the platform? j (b) What impulse did the athlete receive from the platform? j (c) What impulse did the athlete receive from gravity while in contact with the platform? j (d) With what velocity did she leave the platform? j (e) To what height did she jump upon leaving the platform?

Explanation / Answer

Given

mass m = 59 kg

Height h = 0.720 m

time duration is 0 < t <0.8s

F = (9200 N/s)t- (11 500 N/s2)t2

a)

athlet's velocity when she reached the platform is  

mgh = 0.5*mv^2

v= srt(2gh) = sqrt(2*9.8*0.72) m/s

v = 3.76 m/s

b) impulse did the athlete receive from the platform is  

impulse is I = integral (F*t )dt

I = integral ((9200 N/s)t- (11 500 N/s2)t2 )dt

I = 9200*t^2/2 -11500*t^3/3

substituting the value of t =0.8 s we get  

I = 981 N.s

c) impulse did the athlete receive from gravity while in contact with the platform is

I-mg = 981 - 59*9.8*0.8 N.s = 518.44 N.s

d) we know that the impulse is change in momentum

` 981 = mv2 - mv1

v2 = (981+59*3.76)/59 m/s = 20.39 m/s

e) height did she jump upon leaving the platform is h = v2^2/2g = 20.39^2/(2*9.8) m = 21.21 m

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