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Q.
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 $T/3$ seconds?
Motion in a Straight Line
Solution:
$\because h = ut + \frac{1}{2} gt^2 $
$\Rightarrow h = \frac{1}{2} gT^2$
After $\frac{T}{3}$ seconds, the position of the ball,
$ h' = 0 + \frac{1}{2} g(\frac{T}{3})^2 $
$= \frac{1}{2} \times \frac{g}{9} \times T^2$
$ h' = \frac{1}{2} \times \frac{g}{9} \times T^2 = \frac{h}{9}m$ from top
$\therefore $ Position of ball from ground
$ = h - \frac{h}{9} = \frac{8h}{9} m$.
