Question

Two shells have been fired from cannon $A$ and $B$ simultaneously at $t=0$ with velocity as shown in the figure. Find the time when both shells collide in air, given that horizontal distance between cannons is $30\text{m}$.

• A. $5\text{s}$
• B. $2\text{s}$
• C. $3.5\text{s}$
• D. $1.5\text{s}$

Answer 1

The correct option is D $1.5\text{s}$

component of velocity for shell $A$,
$(v_A{)}_x=10\sqrt{3}\times \cos {30}^{\circ }=15\text{m/s}$
$(v_A{)}_y=10\sqrt{3}\times \sin {30}^{\circ }=5\sqrt{3}\text{m/s}$
Similarly, component of velocity for shell $B$,
$(v_B{)}_x=-10\times \cos {60}^{\circ }=-5\text{m/s}$
$(v_B{)}_y=10\times \sin {60}^{\circ }=5\sqrt{3}\text{m/s}$
$\Rightarrow$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 shells will collide

Now, velocity of approach between $A\&B$:
Component of velocity of $A$ w.r.t $B$ in $x-$ direction is
$(v_{AB}{)}_x=(v_A{)}_x-(v_B{)}_x$
$(v_{AB}{)}_x=15-(-5)=20\text{m/s}$
Acceleration of $A$ w.r.t $B$,
$a_{AB}=a_A-a_B=-g-(-g)=0$
$\Rightarrow$Relative separation between shells $A\&B$ in $x-$direction is,
$d=30\text{m}$
$\Rightarrow$Time taken by shells to collide will be:
$t=\frac{d}{(v_{AB}{)}_x}$
$\therefore t=\frac{30}{20}=1.5\text{s}$