Physics
posted 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 impulsemomentum theorem.

Physics 
Kai
haha i think i am in the same physics class with u...

Physics 
Doon
Me too haha. Hope someone solves it soon...

Physics 
ian
kaminsky hooray :p

Physics 
AH
LOL
Welcome to the club guys :P 
Physics 
Phil
hah I love how everyone in physics with Kaminsky is looking here

Physics 
Mike
I don't wanna write it all out so I'll just give you the general formula.
#2: vf=√[2*(m_1gLsinθm_2gL0.5kL^2μkm_1gcosθL)/(m1+m2)]
Just to be clear, all of those values^^^ are under the square root sign.
#4: v1f=[(v1)(m_1m_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. 
Physics 
ian
i cant get the answer... thx for the help tho mike.
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