Physics Question: <span title="Impuse/Momentum/Speed/Height">Impuse/Momentum/Speed/Hei...</span>

Question

Starting from rest, a 65.0 kg athlete jumps down onto the platform from a height of 0.6m While she is in contact with the platform during the time interval 0<t<0.8s, the force she exerts on it is described by the function:

F = 9200t = 11500t^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?

I got (a). That's it :( If possible I'd prefer to have an equation to solve (by myself) and an explanation for why it's there instead of an answer.

Answer 1

You should have gotten impulse I = ∫F(t)*dt.  Impulse = momentum change.  Assuming the diver is stopped at the end of the given time, her momentum change is m*v, so her velocity was I/m.  The platform absorbed all of the kinetic energy she had when she reached the platform (assuming a &quot;lossless&quot; platform spring), and she will have the same energy when she leaves, so her velocity will be the same (KE = 0.5*m*v^2).  Her height above the platform will be the height at which her potential energy equals the kinetic energy, m*g*h = 0.5*m*v^2:  h = 0.5*v^2 / g.