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2.Motion in Straight Line
hard
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
$9 : 4 : 1$
D
$1 : (\sqrt 2 -1): (\sqrt 3 - \sqrt 2)$
Solution
$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$
$\mathrm{h}=0+\frac{1}{2} \mathrm{gt}^{2}$
$t=\sqrt{\frac{2 h}{g}}$
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 $'h'$
$\mathrm{t}_{3}=\sqrt{\frac{6 \mathrm{h}}{\mathrm{g}}}-\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$
$\mathrm{t}_{1}: \mathrm{t}_{2}: \mathrm{t}_{3}=1: \sqrt{2}-1: \sqrt{3}-\sqrt{2}$
Standard 11
Physics
