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Q.

A ball is released from the top of a tower of height $h$ metre. It takes $T$ sec to reach the ground. What is the position of the ball in $T / 3$ sec?

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a

$\frac{h}{9}$ metre from the ground

b

$7 h / 9$ metre from the ground

c

$8 h / 9$ metre from the ground

d

$17 h / 18$ metre from the ground

answer is C.

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Detailed Solution

We have $h = \frac{1}{2} g T^{2}$

In $\frac{T}{3}$ sec, the distance fallen $= \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{h}{9}$

$∴$ Position of the ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9} m$

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A ball is released from the top of a tower of height h metre. It takes T sec to reach the ground. What is the position of the ball in T/3 sec?