Impact force on knee from rollerblading fall at 5 mph

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lettam
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I need help on how to approach this problem and which equations I can use.
It was a problem that we were supposed to think about, but even that's difficult. Here it is:

If there is a 70 lb. girl roller blading at a constant speed of 5 mph and she suddenly hits a rock and trips downward, what would be the impact force on her knee if it is 1 ft above the ground and the distance traveled after impact is 0.25 in?

We had a similar problem without the girl traveling at an initial speed which was simple enough using conservation of energy, potential and kinetic energy, impact velocity, and work. But I'm not sure how the rotational momentum from her fall downward comes into play. Can anyone point me in the right direction? PLEASE!
 

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What do you mean by the distance traveled after impact ?
Any data on the center of mass ?
 
Welcome to PF!

An interesting problem. I would simplify it by assuming that the tripping changes the girl's velocity from horizontal to vertical. So she essentially falls from a height of 1 foot with an initial speed of 5 mph. So what's her KE at the time of impact?

When she hits the ground, the ground exerts a force on her that brings her to rest over a distance of 0.25 inches. So, how much work does that force do? And what's the average force on her knee?
 
Doc Al said:
When she hits the ground, the ground exerts a force on her that brings her to rest over a distance of 0.25 inches. So, how much work does that force do?
Is the force the ground exerts on her knee friction? So the change in KE would be the amount of work done by this friction?
 
I am assuming that the girl falls straight down and sinks into the ground. So the force of the ground on her knee would be a normal force, not friction. But, yes, the change in KE will equal the work done by this normal force.
 
She sinks into the ground!? Ouch. I thought she would slide 0.25 in. across the ground, which is why I asked the question.