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.
haha i think i am in the same physics class with u...
Me too haha. Hope someone solves it soon...
kaminsky hooray :p
LOL
Welcome to the club guys :P
hah I love how everyone in physics with Kaminsky is looking here
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.