A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

Answers (1)

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	imes 9.8	imes 19.6\ v=sqrt2	imes 9.8	imes 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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