Two bodies $\mathrm{A}$ (of mass $1 \mathrm{~kg}$ ) and $\mathrm{B}$ (of mass $3 \mathrm{~kg}$ ) are dropped…
Two bodies $\mathrm{A}$ (of mass $1 \mathrm{~kg}$ ) and $\mathrm{B}$ (of mass $3 \mathrm{~kg}$ ) are dropped from heights of $16 \mathrm{~m}$ and $25 \mathrm{~m}$, respectively. The ratio of the time taken by them to reach the ground is:
$4 / 5$
$5 / 4$
$12 / 5$
$5 / 12$
Solution
We know that time taken by a body from height $h$ to reach the ground is.
$\begin{aligned}
t & =\sqrt{\frac{2 h}{g}} \\
\therefore \quad \frac{t_{\mathrm{A}}}{t_{\mathrm{B}}} & =\frac{\frac{2 h_{\mathrm{A}}}{g}}{\frac{2 h_{\mathrm{B}}}{g}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}} \\
& =\sqrt{\frac{16}{25}}=\frac{4}{5} .
\end{aligned}$