Question

Two particles are projected from the top of a tower with velocities $10\text{m/s}$ and $80\text{m/s}$ in horizontal but in opposite directions. After what time $t$ their velocity vectors will be mutually perpendicular to each other? (Take $g=10{\text{m/s}}^2$)

• A. $2\sqrt{2}\text{sec}$
• B. $2\text{sec}$
• C. $8\text{sec}$
• D. $4\text{sec}$

Answer 1

The correct option is A $2\sqrt{2}\text{sec}$


Let $u_1$ and $u_2$ be the initial velocities of two particles projected from the top of the tower.
${\overrightarrow{u}}_1=80\hat{i}m/s$ and ${\overrightarrow{u}}_2=-10\hat{i}m/s$

Let $t$ be the time after which their velocity vectors are mutually perpendicular.
At time $t$, velocity of both the particles can be given by
${\overrightarrow{v}}_1=80\hat{i}-\left(gt\right)\hat{j}$ and
${\overrightarrow{v}}_2=-10\hat{i}-\left(gt\right)\hat{j}$

For the two velocity vectors to be perpendicular, we can write
${\overrightarrow{v}}_1.{\overrightarrow{v}}_2=0$
$⟹(80\hat{i}-(gt\left)\hat{j}\right).(-10\hat{i}-(gt\left)\hat{j}\right)=0$
$⟹-800+g^2{t}^2=0$
$⟹t=\sqrt{8}=2\sqrt{2}\text{sec}$

Hence option $\left(a\right)$ is the correct answer