Question

Two particles moves in a uniform gravitational field of the Earth. At the initial moment the particles were located at one point and moved with velocities $9m/s$ and $4m/s$ horizontally in opposite direction. The distance between the particle when their velocity are perpendicular to each other will be (Given, $g=10m/s^2$)

• A. $7.8m$
• B. $15.6m$
• C. $36.8m$
• D. $13m$

Answer 1

The correct option is A $7.8m$

Let the particles form angle $\alpha ,\beta$ with the horizontal when their velocities are perpendicular to each other.

The downward velocities are same and equal to $\frac{1}{2}g{t}^2=v$ and $\alpha +\beta ={90}^{\circ }$,

$tan\alpha =cot\beta$

$⟹\frac{v}{9}=\frac{4}{v}$

$⟹v=6m/s$

Thus the time elapsed to achieve this speed $=t$ can be obtained by

$v=gt$

$⟹t=0.6s$

The relative horizontal velocity of the particles=$9m/s+4m/s=13m/s$

Thus the distance between the particles at that moment=$13m/s\times 0.6s=7.8m$