Question

Two particles $A$ and $B$ are projected simultaneously from the towers of same height as shown in the figure with velocities $V_A=20\text{m/s}$ and $V_B=10\text{m/s}$ respectively. They collide in air after 0.5 seconds. Find '$x$'.

• A. $5\text{m}$
• B. $\frac{5}{\sqrt{3}}\text{m}$
• C. $5\sqrt{3}\text{m}$
• D. $10\sqrt{3}\text{m}$

Answer 1

The correct option is C $5\sqrt{3}\text{m}$

Relative velocity of A w.r.t B in horizontal direction, $(u_{rel}{)}_x=u_{Ax}-u_{Bx}=v_A\cos {30}^{\circ }-0=v_A\cos {30}^{\circ }$
Relative accelerartion, $a_{rel}=0$
(No acceleration in horizontal direction)
$\left(S_{rel}\right)=(u_{rel}{)}_{x}t+\frac{1}{2}(a_{rel}{)}_x{t}^2$
$\Rightarrow x=\left(v_A\cos \theta \right)t+0$
$\Rightarrow x=20\cos {30}^{\circ }\left(\frac{1}{2}\right)$
$=20\frac{\sqrt{3}}{2}\times \frac{1}{2}$
$=5\sqrt{3}\text{m}$
So, the correct answer is option (c).