Agolf ball of mass m = 0.21 kg is dropped from a height h. it interacts with the floor for t = 0.115 s, and applies a force of f = 16.5 n to the floor when it elastically collides with it.

write an expression for the ball's velocity, v, just after it rebounds from the floor. (hint: the fact that the collision is elastic is important when solving this problem.) use only t, f, m

Elastic colision => conservation of kinetic energy => constant speed: |V1| =|V2| = V

Momentum before collision: - mV

Momentum after collision:  mV

The signs of the two momenta are different because the direction of the ball is reversed during collision.

Change in momentum: mV - (-mV) = mV + mV = 2mV

Impulse = F*t = Change in momentum

F*t = 2mV

V = F*t / (2m)

That is the expression requested.

Now, to find V, you just have to plug in F, t and m values.