Question

Two particles A and B are shot from the same height at $t=0$ in opposite directions with horizontal velocities $3m/s$ and $4m/s$ respectively. If they are subjected to the same vertical acceleration due to gravity ($g=9.8m/s^2$). the distance between them when their velocity vectors become mutually perpendicular is:

• A. $1.059m$
• B. $1.412m$
• C. $2.474m$
• D. $9.8m$

Answer 1

The correct option is C $2.474m$

Let the velocity of A be $3\hat{i}+u\hat{j}=\overrightarrow{V_A}$

Let the velocity of B be $-4\hat{i}+u\hat{j}=\overrightarrow{V_B}$

where 'u' is the vertical velocity which will be same for both as both are thrown simultaneously. Now,

$\overrightarrow{V_A}.\overrightarrow{V_B}=0(3\hat{i}+u\hat{j})(-4\hat{i}+v\hat{j})=0-12+u^2=0u=\sqrt{12}u=0+9.8tt=\frac{\sqrt{12}}{9.8}s$

Relative velocity of A w.r.t B $=3-(-4)=3+4=7m/s$

Distance $=7\times \frac{\sqrt{1}2}{9.8}=2.474m$