MCQ
A particle is released from rest from a tower of height $3\, h$. The ratio of times to fall  equal heights $h$, i.e., $t_1 : t_2 : t_3$ is
  • A
    $\sqrt 3 :\sqrt 2 :1$
  • B
    $3 : 2 : 1$
  • C
    $9 : 4 : 1$
  • $1 : (\sqrt 2 -1): (\sqrt 3 - \sqrt 2)$

Answer

Correct option: D.
$1 : (\sqrt 2 -1): (\sqrt 3 - \sqrt 2)$
d
$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$\mathrm{h}=0+\frac{1}{2} \mathrm{gt}^{2}$

$t=\sqrt{\frac{2 h}{g}}$

For first height $'h"$

$t_{1}=\sqrt{\frac{2 h}{g}}$

For $II$ height $'h'$

$\mathrm{t}_{2}=\sqrt{\frac{4 \mathrm{h}}{\mathrm{g}}}-\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$

For $III$ height $'h'$

$\mathrm{t}_{3}=\sqrt{\frac{6 \mathrm{h}}{\mathrm{g}}}-\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$

$\mathrm{t}_{1}: \mathrm{t}_{2}: \mathrm{t}_{3}=1: \sqrt{2}-1: \sqrt{3}-\sqrt{2}$

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