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A body is released from the top of a tower ' $\mathrm{H}$ ' metre high. It takes $t$ second to reach the ground. The height of the body $\frac{t}{2}$ second after release is
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The correct answer is:
$3 \frac{\mathrm{H}}{4}$ metre from ground
Let the body be at $\mathrm{x}$ from the top after $\frac{\mathrm{t}}{2} \mathrm{~s}$.
$\begin{aligned}
\therefore \quad \mathrm{x} & =\frac{1}{2} \mathrm{~g} \frac{\mathrm{t}^2}{4}=\frac{\mathrm{gt}^2}{8}.... (i) \\
\mathrm{H} & =\frac{1}{2} \mathrm{gt}^2 .... (ii)
\end{aligned}$
Eliminating $\mathrm{t}$ from (i) and (ii), we get $\frac{8 \mathrm{x}}{\mathrm{g}}=\frac{2 \mathrm{H}}{\mathrm{g}} \Rightarrow \mathrm{x}=\frac{\mathrm{H}}{4}$
$\therefore \quad$ Height of the body from the ground $=\mathrm{H}-\frac{\mathrm{H}}{4}=\frac{3 \mathrm{H}}{4}$ metres
$\begin{aligned}
\therefore \quad \mathrm{x} & =\frac{1}{2} \mathrm{~g} \frac{\mathrm{t}^2}{4}=\frac{\mathrm{gt}^2}{8}.... (i) \\
\mathrm{H} & =\frac{1}{2} \mathrm{gt}^2 .... (ii)
\end{aligned}$
Eliminating $\mathrm{t}$ from (i) and (ii), we get $\frac{8 \mathrm{x}}{\mathrm{g}}=\frac{2 \mathrm{H}}{\mathrm{g}} \Rightarrow \mathrm{x}=\frac{\mathrm{H}}{4}$
$\therefore \quad$ Height of the body from the ground $=\mathrm{H}-\frac{\mathrm{H}}{4}=\frac{3 \mathrm{H}}{4}$ metres
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