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 when their velocities are mutually perpendicular.

• A. $\frac{2\sqrt{u_1{u}_2}}{g}$
• B. $\frac{\sqrt{u_1{u}_2}}{g}$
• C. $\frac{\sqrt{4u_1{u}_2}}{g}$
• D. $\frac{\sqrt{8u_1{u}_2}}{g}$

Answer 1

The correct option is B $\frac{\sqrt{u_1{u}_2}}{g}$


At time $t$,
Velocity of first body
$\overrightarrow{v_1}=-u_1\hat{i}-gt\hat{j}$
Velocity of second body $\overrightarrow{v_2}=u_2\hat{i}-gt\hat{j}$
When velocities are perpendicular,
$\overrightarrow{v_1}.\overrightarrow{v_2}=0\Rightarrow -u_1{u}_2+g^2{t}^2=0\Rightarrow t=\frac{\sqrt{u_1{u}_2}}{g}$