Question

Two bodies are projected from the top of a tower in opposite directions with velocities of $u_1$ and $u_2$ simultaneously. Find the time interval when their
i) velocities are mutually perpendicular
ii) displacements are mutually perpendicular.

• A. (i) $t=\frac{\sqrt{u_1{u}_2}}{g}$; (ii) $t=\frac{2}{g}\sqrt{u_1{u}_2}$
• B. (i) $t=\sqrt{\frac{2u_1{u}_2}{g}}$; (ii) $t=3\sqrt{u_1{u}_2}$
• C. (i) $t=\sqrt{\frac{u_1{u}_2}{2g}}$; (ii) $t=\frac{4}{3}\sqrt{u_1{u}_2}$
• D. (i) $t=\sqrt{\frac{2u_1{u}_2}{3g}}$; (ii) $t=\sqrt{u_1{u}_2}$

Answer 1

The correct option is A (i) $t=\frac{\sqrt{u_1{u}_2}}{g}$; (ii) $t=\frac{2}{g}\sqrt{u_1{u}_2}$


(i) $\overrightarrow{v_1}=-u_1\hat{i}-gt\hat{j}$
$\overrightarrow{v_2}=u_2\hat{i}-gt\hat{j}$
When velocities are perpendicular, $\overrightarrow{v_1}.\overrightarrow{v_2}=0$
$\Rightarrow -u_1{u}_2+g^2{t}^2=0$
$\Rightarrow t=\frac{\sqrt{u_1{u}_2}}{g}$
(ii) $\overrightarrow{S_1}=-u_{1}t\hat{i}+\frac{1}{2}g{t}^2\hat{j}$
$\overrightarrow{S_2}=u_{2}t\hat{i}+\frac{1}{2}g{t}^2\hat{j}$
When displacements are perpendicular i.e. $\overrightarrow{S_1}\perp \overrightarrow{S_2},\overrightarrow{S_1}.\overrightarrow{S_2}=0$
$\Rightarrow -u_1{u}_2{t}^2+\frac{1}{4}{g}^2{t}^4=0$
$\Rightarrow t^2=\frac{4u_1{u}_2}{g^2}\Rightarrow t=\frac{2}{g}\sqrt{u_1{u}_2}$