MCQ
A body dropped from the top of a tower clears $7/16^{th}$ of the total height of the tower in its last second of flight. The time taken by the body to reach the ground is........$s$
- A$2$
- B$3$
- ✓$4$
- D$5$
✓
Answer
Correct option: C.
$4$
c
$\mathrm{H}=\frac{1}{2} \mathrm{gt}^{2}$ $...(1)$
$\mathrm{H}=\frac{1}{2} \mathrm{gt}^{2}$ $...(1)$
$\left(\mathrm{H}-\frac{7}{16} \mathrm{H}\right)=\frac{1}{2} \mathrm{g}(\mathrm{t}-1)^{2}$
$\frac{9}{16} \mathrm{H}=\frac{1}{2} \mathrm{g}(\mathrm{t}-1)^{2}$ $...(2)$
$\mathrm{Eq}^{\mathrm{n}} \cdot(1) /(2)$
$\frac{16}{9}=\frac{\mathrm{t}^{2}}{(\mathrm{t}-1)^{2}}$
$\frac{4}{3}=\frac{\mathrm{t}}{\mathrm{t}-1}$
$\Rightarrow \mathrm{t}=4 \mathrm{sec}$
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