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

Answers (1)

D Deependra Verma

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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