Answer:
13.28m/s
Explanation:
Given
Distance covered by the ball S = 9m
initial velocity u = 0m/s
Required
velocity at the bottom of the hill (final velocity)
Using the equation of motion
v² = u²+2gS
v² = 0²+2(9.8)(9)
v² = 18(9.8)
v² = 176.4
v = √176.4
v = 13.28m/s
Hence the velocity at the bottom of the hill is 13.28m/s