A ball is dropped from rest at a height of 3.7 m. How fast will it be going just prior to hitting the ground?

Answer 1

Let

g = 9.81 m/s² be gravitational acceleration,

h = 3.7 m be the initial height of the ball,

t (unknown) be the time at which the ball hits the ground,

v (to be calculated) be the velocity at the time t.

First calculate the time t (it takes the ball to hit the ground) from the general equation

h = gt²/2

From that express t

t = √(2h/g)

Having the expression for t, calculate v from the general equation

v = gt

= g√(2h/g)

= √(2hg)

Substituting actual numbers

v = √(2×3.7×9.81) = 8.52 m/s