Question

Figure shows two projectiles $A$ and $B$ fired simultaneously. Find the minimum distance between them during their flight.

• A. $10\sqrt{3}\text{m}$
• B. $10\text{m}$
• C. $20\text{m}$
• D. $20\sqrt{3}\text{m}$

Answer 1

The correct option is B $10\text{m}$

Here we analyze the motion of $A$ with respect to $B$. We also assign the sign convention. Let us take $+x-$axis as towards right, $+y-$ axis as up.


The relative motion between two projectiles will be a straight line as the acceleration acting on both is $g$ and so the relative acceleration will be,

$a_{AB}=a_A-a_B=g(-\hat{j})-g(-\hat{j})=0$

Now the relative velocity,
$v_{AB}=v_A-v_B$
$v_A=20\sqrt{3}\cos {60}^{\circ }\left(\hat{i}\right)+20\sqrt{3}\sin {60}^{\circ }(\hat{j}$
$v_B=20\cos {30}^{\circ }(-\hat{i})+20\sin {30}^{\circ }\left(\hat{j}\right)$

So,
$v_{AB}=\left(20\sqrt{3}\cos {60}^{\circ }\right(\hat{i})+20\sqrt{3}\sin {60}^{\circ }(\hat{j})-(20\cos {30}^{\circ }(-\hat{i})+20\sin {30}^{\circ }\left(\hat{j}\right)$
$v_{AB}=(10\sqrt{3}+10\sqrt{3})\hat{i}+(30-10)\hat{j}$
$\Rightarrow v_{AB}=(20\sqrt{3}\hat{i}+20\hat{j})$

Now for finding the angle $\theta$, by the relative velocity direction, we say
$\tan ({60}^{\circ }-\theta )=\frac{20}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow {60}^{\circ }-\theta ={30}^{\circ }$
$\Rightarrow \theta ={30}^{\circ }$

Now, $AB=20\text{m}$ (given)

Hence the shortest distance is $d_{min}=AB\sin {30}^{\circ }=20\sin {30}^{\circ }=20\times \frac{1}{2}=10\text{m}$

Hence option B is the correct answer.