Question

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

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Answer

According to the equation of motion under gravity v2 − u2 = 2gs

Where,

u = Initial velocity of the stone = 0 m/s

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 ms−2

∴ v2 − 02 = 2 × 9.8 × 19.6

⇒ v2 = 2 × 9.8 × 19.6 = (19.6)2

⇒ v = 19.6 ms−1

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

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