Answer: 25.04m/s
Step-by-step explanation:
v^2 = v^2o + 2a(x-xo)
v^2 = velocity
v^2o = initial velocity
a = acceleration
x = final position/distance
xo = initial position/distance
In this case, the initial velocity is 0 since the ball wasn't moving before it was dropped. The final position is 32 as the motion ended after the ball traveled 32m. The initial position is 0. The acceleration is 9.8m/s (free fall). Plug these numbers into the formula:
v^2 = 0 + 2(9.8)(32)
v^2 = 25.04396135
Round to get 25.04m/s