Question

A fixed mortar fires a bomb at an angle of ${53}^{\circ }$ above the horizontal with a muzzle velocity of $80\text{m/s}$. A tank is advancing directly towards the mortar on level ground at a constant speed of $5\text{m/s}$. The initial separation (at the instant mortar is fired) between the mortar and the tank, so that the tank would be hitted is
Take $g=10{\text{m/s}}^2$.

• A. $662.4\text{m}$
• B. $526.3\text{m}$
• C. $486.6\text{m}$
• D. $678.4\text{m}$
  

Answer 1

The correct option is D $678.4\text{m}$


component of velocity for bomb $A$,
$(v_A{)}_x=80\times \cos {53}^{\circ }=48\text{m/s}$
$(v_A{)}_y=80\times \sin {53}^{\circ }=64\text{m/s}$
Similarly, component of velocity for tank $B$,
$(v_B{)}_x=-5\text{m/s}$
$(v_B{)}_y=0$
Component of velocity of $A$ w.r.t $B$ in $x-$ direction is
$(v_{AB}{)}_x=(v_A{)}_x-(v_B{)}_x$
$(v_{AB}{)}_x=48-(-5)=53\text{m/s}$
Component of velocity of $A$ w.r.t $B$ in $y-$ direction is
$(v_{AB}{)}_y=(v_A{)}_y-(v_B{)}_y=64\text{m/s}$
Acceleration of bomb w.r.t tank
$\therefore a_{AB}=a_A-a_B=-g-0=-g{\text{m/s}}^2$

$\Rightarrow$For bomb to hit the tank, the initial separation $\left(d\right)$ between $A\&B$ should be equal to range of bomb $\left(A\right)$ as observed from frame of tank $\left(B\right)$
$\therefore d=\text{Time of flight}\times (v_{AB}{)}_x$
$\Rightarrow d=\frac{2(v_{AB}{)}_y}{g}\times (v_{AB}{)}_x$

$\therefore d=\frac{2\times 64}{10}\times 53=678.4\text{m}$