Answer:
12.5m/s
Step-by-step explanation:
(Assuming the question was asking for the speed just before it hit the ground)
We can use the first key equation of accelerated motion
Vf^2 = Vi^2+2aΔd
Vf^2 = 0 + 2(9.8)(8) (plugged in values, initial velocity is 0 since the ball was at rest)
Vf^2 = 156.8
Vf = 12.5 (squared both sides)