How Is Maximum Force Calculated from Impulse in a Bouncing Ball Scenario?

[merzperson]

Homework Statement

A 200 g ball is dropped from a height of 1.9 m, bounces on a hard floor, and rebounds to a height of 1.5 m. The figure shows the impulse received from the floor.
What maximum force does the floor exert on the ball?

Homework Equations

Kinetics:
x-x_o = v_ot + 0.5at²
v₂ = v₀² + 2a(x - x_o)
v = v_o + at

Momentum and Force:
SF*dt = dP = m*dv

The Attempt at a Solution

What I did was calculate the velocity of the ball just as it hit the ground (-6.10m/s) and just as it rebounded (5.42m/s).
Then, using these I found the dv (5.42 + 6.10 = 11.52m/s) and plugged that into the momentum equation:
dP = m * dv
dP = 0.2 * 11.52 = 2.304 = SFdt

From here I don't know how to calculate the maximum force, as I know virtually nothing about integrals. Also, my work so far is probably wrong somewhere so some help would be greatly appreciated! :)

 

[ahmadmz]

I'm in mechanics and i have not seen a question about max force. But i'll try this. From the graph we can see the Impulse which is the area under the curve is (1/2)(0.005s)F_max
This is equal to the change in momentum which we can calculate.

So F_max = 2*(p2-p1)/0.005

Where p2-p1 = m(v2-v1)