How High Does the Force Platform Indicate the Athlete Jumped?

[szimmy]

Homework Statement

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

Homework Equations

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

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.

 

[gneill]

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.

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

 

[szimmy]

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

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!

 

[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.

 

[gneill]

I just have one more question. Our school doesn't offer calculus based physics (I'm in AP Physics B, but also calc) so we aren't required to do calculus to solve any problems. This normally isn't a problem, but the F versus T graph is a curve is it not? Is it possible for me to find the area under the curve to find the impulse without using calculus? This has me confused now.

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.