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Question 14

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

Answer

 Initial velocity of the stone = u = 0

 Final velocity of the stone = v

 Height of the stone = s = 19.6 m

 Acceleration due to gravity = g = 9.8 m s−2

According to the equation of motion under gravity:

v2 − u2 = 2 gs

v2 − 02 = 2 × 9.8 × 19.6

v2 = 2 × 9.8 × 19.6 = (19.6)2

v = 19.6 m s−1

Hence, the velocity of the stone just before touching the ground is 19.6 m s−1.

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