Answer:
8.29m/s
Explanation:
Given
Height of hill H = 3.50m
Initial velocity u = 0m/s
Required
Final velocity v
Using the equation of motion
v² = u²+2gH
v² = 0²+2(9.81)(3.5)
v² = 0+7(9.81)
v² = 68.67
v = √68.67
v = 8.29m/s
Hence the final velocity of the hoop is 8.29m/s
