Impulse and Energy Question

1. Dec 12, 2012

szimmy

1. The problem statement, all variables and given/known data
The "force platform" is a tool that is used to analyze the performance of athletes by measuring the vertical force as a function of time that the athlete exerts on the ground in performing various activities. A simplified force vs. time graph for an athlete performing a standing high jump is shown in Figure P6.66. The vertical divisions of the graph represent 0.65 kN and the horizontal divisions represent 0.50 s. The athlete started the jump at t = 0.0 s. How high did this athlete jump?
Picture of graph: http://www.webassign.net/sercp/p6-66alt.gif

2. Relevant equations
J = FΔt
J = ΔP
J = m(vf - vi)
FΔt = mvf (since the jump starts from rest)

3. The attempt at a solution
From the graph I was able to get that the impulse is 650N. The problem is I have no idea how to use this to get the height he jumped, I feel like I need a mass to solve this.
Given that J = m(vf - vi)
This means that FΔt / m = vf (vi is 0 because starting from rest)
This vf would be the vi for the jump, so mgh = 1/2 m v^2
h = v^2/2g
h = (650/m)^2/2g
h = 422500 / 2*g*m^2
I'm stuck from here, none of my classmates could figure it out either. If somebody could give me a push in the right direction I would appreciate it. I feel like the solution is obvious and I'm just thinking too hard.

Last edited: Dec 12, 2012
2. Dec 12, 2012

Staff: Mentor

What value does the graph show for time $t \le 0$? Why do you suppose that is?

3. Dec 12, 2012

szimmy

Wow, I can't believe I didn't realize that before. He's standing still so the 650N when he's standing still is mg. Thanks for your help!

4. Dec 12, 2012

szimmy

Disregard this post. I asked a question and then went and tinkered with some stuff and (somehow) made it work. I couldn't find a way to delete the post so I'm just editing it.

Last edited: Dec 12, 2012
5. Dec 12, 2012

Staff: Mentor

Well, since you aren't given an equation for the curve, calculus isn't going to be much help anyways. You're left with "counting squares" or assembling a number of geometric shapes to cover the area to reach an approximation.