Question 14:
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s−2
∴ v2 − o2 = 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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