A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.
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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Welcome to the NCERT Solutions for Class 9 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 6 , Question 14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.....
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