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9. GRAVITATION
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A stone is released from the top of a tower of height $19.6\, m$. Calculate its final velocity just before touching the ground.
A
$19.6\, ms^{-1}$
B
$15.8\, ms^{-1}$
C
$12.7\, ms^{-1}$
D
$13.5\, ms^{-1}$
Solution
According to the equation of motion under gravity $v^2 – u^2 = 2gs$
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 $= 9.8\, ms^{-2}$
$\therefore $ $v^2 – 0^2 = 2 \times 9.8 \times 19.6 $
$v^2 = 2 \times 9.8 \times 19.6 = (19.6)^2$
$v = 19.6 \,ms^{-1}$
Hence, the velocity of the stone just before touching the ground is $19.6\, ms^{-1}$
Standard 9
Science
