Asked by AH

A 0.4 kg ball strikes a smooth level floor with a velocity of 5.0 m/s, 60 degrees below the horizontal.  It rebounds off the floor with an unknown velocity and reaches a maximum rebound height of 0.60 m. If the duration of the collision with the floor is 50 milliseconds, then what was the average (magnitude of the) normal force (in Newtons) the ground applied to the ball?

Hint: Use the impulse-momentum theorem.
Answered by Kai
haha i think i am in the same physics class with u...
Answered by Doon
Me too haha. Hope someone solves it soon...
Answered by ian
kaminsky hooray :p
Answered by AH
LOL

Welcome to the club guys :P
Answered by Phil
hah I love how everyone in physics with Kaminsky is looking here
Answered by Mike
I don't wanna write it all out so I'll just give you the general formula.

#2: vf=√[2*(m_1gLsinθ-m_2gL-0.5kL^2-μkm_1gcosθL)/(m1+m2)]
Just to be clear, all of those values^^^ are under the square root sign.

#4: v1f=[(v1)(m_1-m_2)/(m_1+m_2)]+[(v2)2m_2/(m_1+m_2)]
I'm pretty sure this formula was in this notes, but make sure your answer is positive.
Answered by ian
i cant get the answer... thx for the help tho mike.
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