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.
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.