A stone is released from the top of a tower o | vedclass

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Question

(A) According to the equation of motion under gravity $v^2 - u^2 = 2gs$.Where,$u = 	ext{Initial velocity of the stone} = 0\, ms^{-1}$.$v = 	ext{Fina

Vedclass · Accepted Answer

$19.6$

Vedclass · Answer

(A) According to the equation of motion under gravity $v^2 - u^2 = 2gs$.Where,$u = 	ext{Initial velocity of the stone} = 0\, ms^{-1}$.$v = 	ext{Final velocity of the stone}$.$s = 	ext{Height of the tower} = 19.6\, m$.$g = 	ext{Acceleration due to gravity} = 9.8\, ms^{-2}$.Substituting the values in the equation:$v^2 - 0^2 = 2 	imes 9.8 	imes 19.6$.$v^2 = 19.6 	imes 19.6 = (19.6)^2$.$v = 19.6\, ms^{-1}$.Hence,the velocity of the stone just before touching the ground is $19.6\, ms^{-1}$.

$A$ stone is released from the top of a tower of height $19.6\, m$. Calculate its final velocity just before touching the ground. (in $, ms^{-1}$)

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