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 0.770 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?

(b) What impulse did the athlete receive from the platform?

(c) What impulse did the athlete receive from gravity while in contact with the platform?

(d) With what velocity did she leave the platform?

(e) To what height did she jump upon leaving the platform?

Explanation / Answer

v = Sqrt [2 g h] = sqrt [ 2 * 9.8 * 0.770 ]

= 3.88 m/s, velocity = -3.88 j

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F = (9 200 N/s)t - (11 500 N/s2)t2

(9 200 N/s)t2/2 - (11 500 N/s2)t3/3 ] 0 to 0.8

= 981.33 N.s

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F = m g = 59.0 * (-9.8) = -578.2 N

F * h = -578.2 * 0.77 = -445 N.s

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Ft = 981.33 - 445

Ft = 536.33 N.s [upward ]

impulse will equal a change in momentum Ft = m * delta v

Ft = m (vf - vi)

(Ft / m) + vi = vf

vf = (536.33 / 59) - 3.88

vf = 5.21 m/s [upward]

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h = (vf)2 / 2 g

= 1.38 m

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