A 0.540-kg ball is dropped from rest at a point 2.30 m above the floor. The ball rebounds straight upward to a height of 1.10 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?

Answer 1

first find the velocities at the floor:
1/2 .540*Vi^2=.540*9.8*2.3 and
1/2 .540*vf^2=.540*9.8*1.1

for vi=-sqrt(2*9.8*2.3) and
vf= sqrt(2*9.8*1.1)

impulse = change mmentum
= mass*(vf-vi)=mass*sqrt(19.8)( sqrt(19.9*1.1) sqrt(1.1+2.2)