MHT CET · Physics · Motion In One Dimension
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
- A \(\frac{\mathrm{H}}{2}\) metre from ground
- B \(\frac{\mathrm{H}}{4}\) metre from ground
- C \(3 \frac{\mathrm{H}}{4}\) metre from ground
- D \(\frac{\mathrm{H}}{6}\) metre from ground
Answer & Solution
Correct Answer
(C) \(3 \frac{\mathrm{H}}{4}\) metre from ground
Step-by-step Solution
Detailed explanation
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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