Question

A 1.5-kg playground ball is moving with a velocity of 3.0 m/s directed 30â° below the horizontal just before it strikes a horizontal surface. the ball leaves this surface 0.50 s later with a velocity of 2.0 m/s directed 60â° above the horizontal. what is the magnitude of the average resultant force on the ball?

Answer 1

By components you calculate change in momentum for each component; 
Fx = dPx/dt = (m/t)[VfCos(60) - ViCos(30)] 
Fy = dPy/dt = (m/t)[VfSin(60) - - ViSin(30)] 

Once you get those numbers for Fy & Fx then the magnitude of the net force is; 
F = SqRt[Fx^2 + Fy^2]