If a ball rolls off a cliff that is 40.0 meters tall and lands 10.0 meters from the base of the cliff, with what initial horizontal velocity did the ball roll off the cliff? neglect air resistance. a.2.80 m/s b.3.00 m/s c.3.50 m/s d.4.20 m/s

Option (c) is correct.

Explanation:

initial velocity along vertical=0

velocity along horizontal Vx

acceleration along the vertical=9.8 m/s²

displacement along vertical,h=40 m

distance along horizontal= 10 m

using the kinematic equation

h= vi*t + 1/2 g t²

40=0(t) + 1/2 (9.8)t²

t²=8.163

t=2.857 s

now the displacement along vertical= x= Vx* t

10=Vx (2.857)

Vx=3.5 m/s