# A stone is released from rest from the edge of a building roof 190 m above the ground. Neglecting air resistance, what is the speed of the stone, just before striking the ground.

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The stone is released from rest at the edge of the roof of a building that is 190 m above the ground. The initial velocity of the stone is 0 m/s. When released, the stone is accelerated downwards due to the gravitational force of attraction of the Earth at 9.8 m/s^2 in a direction that is vertically downwards.

Use the formula v^2 - u^2 = 2*a*s where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled. Substituting the values provided in the problem:

`v^2 - 0^2 = 2*9.8*190`

`=> v = sqrt(3724)`

`~~ 61.02 m/s`

The velocity of the stone just before it strikes the ground is approximately 61.02 m/s in a direction vertically downwards.