# Q. 14.     A stone is released from the top of a tower of height $19.6\; m.$ Calculate its final velocity just before touching the ground.

Initial velocity u = 0

Acceleration, a = g = 9.8 m s-2

Distance travelled, s = 19.6 m

Let the final velocity be v

According to the third equation of motion

$\\v^{2}-u^{2}=2as\\ v^{2}-0^{2}=2\times 9.8\times 19.6\\ v=\sqrt{2\times 9.8\times 19.6}\\ v=19.6\ m\ s^{-1}$

Its final velocity just before touching the ground will be 19.6 m s-1.

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