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(C) For a ball released from rest,the total height $h$ covered in time $T$ is given by the equation of motion:$h = ut + \frac{1}{2}gT^2$Since the init

Vedclass · Accepted Answer

$\frac{8h}{9}$ meters from the ground

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(C) For a ball released from rest,the total height $h$ covered in time $T$ is given by the equation of motion:$h = ut + \frac{1}{2}gT^2$Since the initial velocity $u = 0$,we have:$h = \frac{1}{2}gT^2$  --- $(1)$After time $t = \frac{T}{3}$,the distance $h'$ covered by the ball from the top is:$h' = \frac{1}{2}g\left(\frac{T}{3}ight)^2 = \frac{1}{2}g\left(\frac{T^2}{9}ight) = \frac{1}{9} \left(\frac{1}{2}gT^2ight)$Substituting equation $(1)$ into this expression:$h' = \frac{h}{9}$The position of the ball from the ground is the total height minus the distance covered from the top:$	ext{Position from ground} = h - h' = h - \frac{h}{9} = \frac{8h}{9} 	ext{ meters}$.

$A$ ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball in $\frac{T}{3}$ seconds?

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