Question

Two particles move in a uniform gravitational field with an acceleration $g$. At the initial moment the particles were located at one point and move with velocities $v_1=3.0m/s$ and $v_2=4.0m/s$ horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

• A. $7\sqrt{3}m$
• B. $\frac{7\sqrt{3}}{5}m$
• C. $5m$
• D. $\frac{7}{2}m$

Answer 1

The correct option is B $\frac{7\sqrt{3}}{5}m$

Draw a diagram of given situation.

After a time $t$,

$1^{st}$ particle velocity, $v_y=u-gt=4\hat{i}-gt\hat{j}$
$2^{nd}$ particle velocity, $v_y=u-gt=-3\hat{i}-gt\hat{j}$

Particles move perpendicular to each other at time $t$.
$(4\hat{i}-gt\hat{j}).(-3\hat{i}-gt\hat{j})=0$

$\Rightarrow -12+g^2{t}^2=0$

$\Rightarrow t=\sqrt{\left(\frac{12}{100}\right)}\Rightarrow t=\frac{\sqrt{3}}{5}$

Since both the particles had zero initial vertical velocity, distance between the two particles after time $t$ will be horizontal distance.

Distance, $d=3t+4t=7t=7\times \frac{\sqrt{3}}{5}=\frac{7\sqrt{3}}{5}m$