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:

*v _{2} − u_{2} = 2 gs*

v_{2} − 02 = 2 × 9.8 × 19.6

v_{2} = 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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- NCERT Chapter

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