Question

Two bombs have been fired simultaneously from cannon $A$ and $B$, which are similar and have a fixed barrel inclination of ${45}^{\circ }$. The cannons are facing each other and have a horizontal separation of $400\text{m}$, bombs are fired towards each other at a speed of $20\sqrt{2}\text{m/s}$. Find the time when both shells collide in air.

• A. $10\text{s}$
• B. $5\text{s}$
• C. $20\text{s}$
• D. $15\text{s}$

Answer 1

The correct option is A $10\text{s}$

component of velocity for bomb $A$,
$(v_A{)}_x=20\sqrt{2}\times \cos {45}^{\circ }=20\text{m/s}$
$(v_A{)}_y=20\sqrt{2}\times \sin {45}^{\circ }=20\text{m/s}$
Similarly,component of velocity for shell $B$,
$(v_B{)}_x=-(20\sqrt{2}\times \cos {45}^{\circ })=-20\text{m/s}$
$(v_B{)}_y=20\sqrt{2}\times \sin {45}^{\circ }=20\text{m/s}$

$\Rightarrow$Relative velocity component in $y-$ direction for $A\&B$,
Component of velocity of $A$ w.r.t $B$ in $y-$ direction is
$(v_{AB}{)}_y=(v_A{)}_y-(v_B{)}_y=0...(i)$
Hence bombs will collide

Now, velocity of approach between $A\&B$:
$\Rightarrow$ Component of velocity of $A$ w.r.t $B$ in $x-$ direction is
$(v_{AB}{)}_x=(v_A{)}_x-(v_B{)}_x$
$(v_{AB}{)}_x=20-(-20)=40\text{m/s}$
Relative acceleration $a_{AB}=a_A-a_B=-g-(-g)=0$
$\Rightarrow$Relative separation between bombs $A\&B$ in $x-$direction is,
$d=400\text{m}$
$\Rightarrow$Time taken by bombs to collide will be:
$t=\frac{d}{(v_{AB}{)}_x}$
$\therefore t=\frac{400}{40}=10\text{s}$