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A ball is dropped from 21.0 m. How fast is it going just before it hits the ground?

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Answer: 20.29m/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 21 as the motion ended after the ball traveled 21m. 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)(21)

v^2 = 20.28792744

Round to get 20.29m/s

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