Question

A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal heights h is:

• A. $t_1:t_2:t_3$ = 3 : 2 : 1
• B. $t_1:t_2:t_3$ = $1:(\sqrt{2}-1):(\sqrt{3}-2)$
• C. $t_1:t_2:t_3$ = $\sqrt{3}:sqrt2:1$
• D. $t_1:t_2:t_3$ = $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Answer 1

The correct option is D $t_1:t_2:t_3$ = $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

For $A$ to $B-$

$h=\left(a\right)\left(t_1\right)+\frac{1}{2}g{t}_1^2\left[2^{nd}\text{equation of motion.}\right]$

$t_1=\sqrt{\frac{2h}{g}}$..................(1)

From $A$ to $C-$

$2h=\left(0\right)\left(t_2\right)+\frac{1}{2}g{t}_2^2$

$t_2=\sqrt{\frac{4h}{g}}$ ..................(2)

Similarly, $t_3=\sqrt{\frac{6h}{g}}$ ........(3

)

Times taken from $A$ to $B=t_1=\sqrt{\frac{2h}{g}}$

Times taken from $B$ to $C=t_2-t_1=\sqrt{\frac{4h}{g}}-\sqrt{\frac{2h}{g}}=\sqrt{\frac{2h}{g}}(\sqrt{2}-1)$

Time taken from $C$ to $D=t_3-t_2=\sqrt{\frac{6h}{g}}-\sqrt{\frac{4h}{g}}=\sqrt{\frac{2h}{g}}(\sqrt{3}-\sqrt{2})$

$\therefore$ Ration $=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Option- D