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 60.0-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

where F is in newtons and t is in seconds. Help solve parts c and d please

Explanation / Answer

The force F by Newton's Second Law is F = dP/dt, or The net force on an object is the time rate of change of the momentum of the object.

Over the eight tenths of a second she is on the platform,

F = dP/dt = 9200N/s t - 11500 N/s^2 t^2
so

dP = Fdt = 9200N/s^2 t - 11500n/s^2 t^2

Integrating wrt t on the boundaries of 0 s to 0.8s will give the impulse she gives the table (N3 says this is equal in magnitude and opposite in direction to the impulse the table gives her)

integrating, we get

J = [4600N/s*t^2 - 3833N/s^2 * t^3] on boundaries 0 to 0.8s, so the impulse is (in magnitude), by evaluating the function at tf= 0.8s and ti = 0 we get the answer to (a)

(a) J = 981 Ns

The direction of the impulse the platform gives her is obviously upward because it has changed her velocity from downward to upward.

Now for (b): To get her speed as she reaches the platform, just use your equations of motion in the y direction. She falls from a height of 0.690 m, so her speed after this (starting from rest) is

V^2 = Vo^2 + 2*g*(yo-y), where g = 9.8m/s^2, Vo = 0, and (yo-y) = 0.690 m

so
V^2 = 2*9.8*0.690 m^2/s^2 = 13.5 m^2/s^2

and thus

(b) V = 3.68 m/s

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

Po = 67.0kg * V = 67.0kg * 3.68 m/s = 246 kgm/s = 246 Ns

so letting positive be upward her initial momentum is

Po = -246 Ns

Her final momentum after she leaves the table is given by

Pf - Po = J

so

Pf = Po + J = -246 Ns + 981 Ns = 735 Ns

So her speed upward when she leaves the platform is

(c) Vf = Pf/m = 735Ns/67.0kg = 10.97 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 = (10.97m/s)^2/(2*9.8m/s^2)

(d) h = 6.14 m!!!!!!!!!!!!! Wow. I would say that is impossible. But I cannot see my error.

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