Question

Two particles are projected from the top of a tower with velocities $20m/s$ and $80m/s$ in horizontal but in opposite directions. After what time $t$ their velocity vectors will be mutually perpendicular to each other? (Take $g=10m/s^2$)

• A. $t=2$ sec
• B. $t=4$ sec
• C. $t=6$ sec
• D. $t=8$ sec

Answer 1

The correct option is B $t=4$ 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 ${\overrightarrow{u}}_2=-20\hat{i}m/s$
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
${\overrightarrow{v}}_2=-20\hat{i}-\left(gt\right)\hat{j}$
For the two velocity vectors to be perpendicular, we can write
${\overrightarrow{v}}_1.{\overrightarrow{v}}_2=0$
$⟹(80\hat{i}-(gt\left)\hat{j}\right).(-20\hat{i}-(gt\left)\hat{j}\right)=0$
$⟹-1600+g^2{t}^2=0$
$⟹t=4sec$