Question

An object is dropped from rest from a large height. Assume $g$ to be constant throughout the motion. The time taken by it to fall through successive distances of $1\text{m}$ each will be

• A. same, which is $\sqrt{\frac{2}{8}}$ seconds
• B. in the ratio of the square roots of the integers 1, 2, 3, …
• C. in the ratio of the difference in the square roots of the integers, i.e., $\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})$, ...
• D. in the ratio of the reciprocals of the square roots of the integers, i.e., $\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},$ ...

Answer 1

The correct option is C in the ratio of the difference in the square roots of the integers, i.e., $\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})$, ...

For 1st $1m$:
as $u=0$
$s=\frac{1}{2}g(t_1{)}^2$
$1=5(t_1{)}^2$
$t_1=\frac{1}{\sqrt{5}}$

For 2nd $1m$:
We need subtract the time taken by body to fall $1m$ from the total time taken by the body to fall $2m$.
$\therefore 2=5T^2$, here $T=$ total time
$T=\sqrt{\frac{2}{5}}$
$t_2=T-t_1=\sqrt{\frac{2}{5}}-\sqrt{\frac{1}{5}}$

Similarly,
for nth $1m$:
$t_n=\sqrt{\frac{n}{5}}-\sqrt{\frac{n-1}{5}}$

So the ratio will be:
$t_1:t_2:t_3:...:t_n=\sqrt{\frac{1}{5}}:\sqrt{\frac{2}{5}}-\sqrt{\frac{1}{5}}:\sqrt{\frac{3}{5}}-\sqrt{\frac{2}{5}}:...$
$t_1:t_2:t_3:...:t_n=\sqrt{1}:\sqrt{2}-\sqrt{1}:\sqrt{3}-\sqrt{2}:...$