Assuming zero drag, how long, in seconds, would it take a super ball to fall 25 feet? What would its velocity be at that point (in mph)?

Answer 1

Use the kinematics equation:

Δy = viΔt + (1/2)gΔt²

Here
Δy=-25'
vi=0 (initial velocity)
g=-32.2 ft/s²

Δt=√(2Δy/g)
=√(2*25/32.2)
=1.246 s. (approx.)

Velocity
=1.246*(-32.2) fps
=40.12 fps
=40.12/5280*3600 mph
=27.4 mph (approx.)