Analyzing an Athlete's Performance Using a Force Platform

[Zero Gravity]

Good day, I have a homework problem I'd like help with. I got into getting the impulse (not sure if correct, though), but don't really know how to get the rest of the things. Any tips on what direction I should be going are appreciated.

Homework Statement

Starting from rest, 65kg athlete jumps down onto a platform that is .600m high. While the athlete is on contact with the platform during the time interval $$0 < t < 0.800 s$$, the force exerted is discribed by the function $$F = (9200 N/s)t - (11500 N/s^2)t^2$$.

$$m = 65 kg$$
$$h_p = .600 m$$(platform)
$$V_0 = ?$$ (Is this supposed to be 0? She starts from rest but jumps up and then goes down to the platform)
$$V_f = ?$$
$$h_m = ?$$(max)

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

(b)With what speed did she reach the platform?

(c)With what speed did she leave it?

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

Homework Equations

$$F = (9200 N/s)t - (11500 N/s^2)t^2$$

$$\vec{I} \equiv \int_t^t \Sigma \vec{F}dt$$(top t is final, bottom initial...)

The Attempt at a Solution

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

I punched in $$F = (9200 N/s)t - (11500 N/s^2)t^2$$ values into $$\vec{I} \equiv \int_t^t \Sigma \vec{F}dt$$

$$\vec{I} \equiv \int_0^.8 (9200 N/s)(t) - (11500 N/s^2)(t^2)$$

$$\equiv \int (4600 N) (.8 s)^2 - (3833.33 N) (.8 s)^3$$

$$\equiv (2944 N/s) - (1962.5 N/s)$$

$$\vec{I} \equiv 981.5 N/s$$

 

[Delphi51]

Your work for (a) looks good.
The question isn't entirely clear, but it must mean she comes to a stop at the end of that 0.8 s interval. If so, you can use impulse = m*Δv to do (b).

 

[Zero Gravity]

Your work for (a) looks good.
The question isn't entirely clear, but it must mean she comes to a stop at the end of that 0.8 s interval. If so, you can use impulse = m*Δv to do (b).

Using $$I = m*\Delta v$$

I get $$\Delta V = 15.1 m/s$$

$$\frac {I}{m} = \Delta v$$

$$\frac {981.5 N/s}{65 kg} \equiv \Delta v$$

$$\Delta v \equiv 15.1 m/s$$

So Delta V would be the landing speed since when she starts falling down the velocity would be 0. Or would $$\Delta v = 15.1 m/s$$ be the difference in velocity of when she lands and when she jumps(which means it would count the time she ascends as well)?$$V_f^2 = V_i^2 +2a(Y_f-Y_i)$$
vf ^2= vi^2 +2a (yf-yi)
15.1^2 = 0 + 19.62 (yi)
228.01/19.62=y
y=11.62m

Falling from a distance of 11m is kinda impossible isn't it? So 15.1m/s is the difference from when she jump... right?

 

[Delphi51]

There is no mention of a jump in the question.
If she does jump, how do we know what part of the impulse stops the fall and what part makes the jump? Maybe you should write out the whole question - perhaps we are missing some more information.

 

[Zero Gravity]

There is no mention of a jump in the question.
If she does jump, how do we know what part of the impulse stops the fall and what part makes the jump? Maybe you should write out the whole question - perhaps we are missing some more information.

A force platform is a tool used to analyze the performance of athletes by measuring the vertical force that the athlete exerts on the ground as a function of time.Starting from rest, a 65kg athlete jumps down onto the platform from a height of 0.600m. While she is in contact with the platform during the time interval 0 <t <.800 s, the force she exerts on it is described by the function
F = (9200 N/s)t - (11500 N/s^2)t^2

a)What impulse did the athlete receive from the platform?
b)With what speed did she reach the platform?
c)With what speed did she leave it?
d)To what height did she jump upon leaving the platform?

Is what's on the book. I too, find it not to make much sense. :(