Question

Two guns $A$ and $B$ can fire bullets at speed of $3\text{km/s}$ and $4\text{km/s}$ respectively. , They are fired in all possible directions, from a point on the horizontal ground. The ratio of maximum area covered by the fired bullets on the ground will be:

• A. $\frac{3}{4}$
  
• B. $\frac{9}{16}$
  
• C. $\frac{81}{256}$
  
• D. $1$

Answer 1

The correct option is C $\frac{81}{256}$


When the guns fire bullets at some angle $\theta$,it will follow the prabolic path for a projectile.
Range of the projectile is given by:
$R=\frac{u^2\sin 2\theta }{g}.......\left(1\right)$
$\therefore$ Maximum area of fired bullets will correspond to area of circle with radius equal to maximum range ($r=R_{\text{max}}$).
$\Rightarrow$Range will be maximum for $\sin 2\theta =1$ i.e $\theta ={45}^{\circ }$
$i.e,R_{\text{max}}=\frac{u^2}{g}$

Ratio of maximum area;
$\Rightarrow \frac{{\text{(Maximum Area)}}_A}{{\text{(Maximum Area)}}_B}=\frac{\pi (R_{\text{max}}{)}_A^2}{\pi (R_{\text{max}}{)}_B^2}=\frac{\frac{u_A^4}{g^2}}{\frac{u_B^4}{g^2}}=\frac{u_A^4}{u_B^4}$

$\therefore \frac{{\text{(Maximum Area)}}_A}{{\text{(Maximum Area)}}_B}=\frac{(3{)}^4}{(4{)}^4}=\frac{81}{256}$