# A stone is dropped from the top of a building that is 320 m high. What is the velocity of the stone when it reaches a height of 180 m.

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A stone is dropped from a height of 320 m. This indicates that its initial velocity is 0. Due to the gravitational pull of the Earth the stone is accelerated downwards at 9.8 m/s^2. The velocity of the stone at a height of 180 m has to be determined. When the height of the stone is 180 m it has traveled downwards a distance equal to 320 - 180 = 140 m.

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 of the body and s is the distance traveled. Substituting the values given v^2 - 0^2 = 2*9.8*140

=> v = `sqrt(2744) ~~ 52.38` m/s^2

**The velocity of the stone at a height of 180 m is approximately 52.38 m/s^2 in a direction vertically downwards.**