Question

A particle is released from rest from a tower of height $3h$. 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. $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Answer 1

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

The particle will travel equal distance in successive time intervals $t_1,t_2\text{and}{t}_3,$ if
$h=\frac{1}{2}g{t}_1^2\dots \left(i\right)$
$2h=\frac{1}{2}g(t_1+t_2{)}^2\dots (ii)$
$3h=\frac{1}{2}g(t_1+t_2+t_3{)}^2\dots (iii)$
Put $h$ from equation (i) in equation (ii) we get
$2\left(\frac{1}{2}g{t}_1^2\right)=\frac{1}{2}g(t_1+t_2{)}^2$
$2t_1^2=(t_1+t_2{)}^2$
$\sqrt{2}{t}_1-t_1=t_2$
$\Rightarrow (\sqrt{2}-1)=\frac{t_2}{t_1}\dots \left(iv\right)$
Put $h$ from equation (i) in equation (iii) we get
$3\left(\frac{1}{2}g{t}_1^2\right)=\frac{1}{2}g(t_1+t_2+t_3{)}^2$
$3t_1^2=(t_1+t_2+t_3{)}^2$
$\sqrt{3}{t}_1-t_1=t_2+t_3$
$(\sqrt{3}-1)=\frac{t_2}{t_1}+\frac{t_3}{t_1}$
By using equation (iv)
$\sqrt{3}-1=\sqrt{2}-1+\frac{t_3}{t_1}$
$\sqrt{3}-\sqrt{2}=\frac{t_3}{t_1}\dots \left(v\right)$
By using equation (iv) and (v)
or $t_1:t_2:t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$