Question

In the figure shown, the two projectiles are fired simultaneously.
The minimum distance between them during their fight is

• A. 20 m
• B. 30 m
• C. 10 m
• D. zero

Answer 1

The correct option is A 20 m

Resolve the velocity of particle at point $A$ and $B$ in x-direction and y-direction.

lets $V_{Ax}=$ velocity component of $A$ particle along X-axis

$V_{Ay}=$ velocity component of $A$ along y-axis.

$V_{Ax}=20\sqrt{3}cos60=10\sqrt{3}m/s$

$V_{Ay}=20\sqrt{3}sin60=30m/s$

$V_{Bx}=-20cos30=-10\sqrt{3}m/s$

$V_{By}=20sin30=10m/s$

Lets consider the particle $B$ is at rest, then the relative velocity of particle A with respect to B is

$V_{AB}=(V_{Ax}-V_{Bx})\hat{i}+(V_{Ay}-V_{By})\hat{j}$

$V_{AB}=20\sqrt{3}\hat{i}+20\hat{j}$. . . . .(1)

Now, new vector diagram is as shown in the fig.

The particle A moves with a velocity of $V_{AB}$ make angle $\theta$

$tan\theta =\frac{V_{AB\left(y\right)}}{V_{AB\left(x\right)}}=\frac{20}{20\sqrt{3}}$

$tan\theta =\frac{1}{\sqrt{3}}$

$\theta ={30}^0$

The distance between two particle is $20\sqrt{3}$.

As we know

$sin{30}^0=\frac{d}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$,

$d=20m$

The minimum distance beteen them during their flight is $20m$

The correct option is A.