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 $v_{1} = 3.0\, m/s$ and $v_{2} = 4.0\, m/s$ horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

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

Answer 1

Answer: (C)

Let the particles move perpendicular to each time $t$.

Hence $\left(4\hat{i}-gt\hat{j}\right).\left(-3\hat{j}-gt\hat{j}\right)=0 $
$\Rightarrow -12 + g^{2}t^{2}=0 $
$\Rightarrow t=\sqrt{\left(\frac{12}{1000}\right)}\Rightarrow t=\frac{\sqrt{3}}{5}$
Hence distance,
$d=\sqrt{\left(ut+3t^{2}\right)+\left(\left(h-\frac{1}{2}gt^{2}\right)-\left(h-\frac{1}{2}gt^{2}\right)\right)^{2}}$
$=7t=\frac{7\sqrt{3}}{5} m$