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 67.2-kg athlete jumps down onto the platform from a height of 0.606 m. While she is in contact with the platform during the time interval 0 < t < 0.800 s, the force she exerts on it is described by the function F = 9 200t 11 500t2 where F is in newtons and t is in seconds.

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

(b) With what speed did she reach the platform?
____________ m/s

(c) With what speed did she leave it?
_____________ m/s

(d) To what height did she jump upon leaving the platform?
_______________ m

Explanation / Answer

a)

Force varies in the time interval from 0 to .8 s. Find the area under the curve of force vs time

Impulse = Force x Time

(9200t 11 500t2) dt. Evaluated it from 0 to .8

= 981 N-s

b)

v = sqrt(2*a*h) = sqrt(2*9.8*.606) =3.44 m/s

Now

for (c). Her impulse given to her by the platform changes her momentum. Initially her momentum is downward and is

Po = 67.2kg * V = 67.2kg * 3.44 m/s = 231.1 kgm/s = 231.1 Ns

so letting positive be upward her initial momentum is

Po = -231.1Ns

Her final momentum after she leaves the table is given by

Pf - Po = J

so

Pf = Po + J = -231.1 Ns + 981 Ns = 749.9 Ns

So her speed upward when she leaves the platform is

(c) Vf = Pf/m = 749.9Ns/67.2kg =11.15 m/s


(d) She will attain a height with that initial speed again given by

V^2 = Vf^2 - 2g(y-yo)

where V = 0 (the top of her jump)

and y-yo her height h

0 = Vf^2 - 2gh

2gh = Vf^2

h = Vf^2/2g = (11.15m/s)^2/(2*9.8m/s^2)

(d) h = 6.34 m

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