Answer
4
Work Step by Step
Let's take,
The initial horizontal velocity of bullets $=V$
The initial vertical velocity of bullets $=0$
The horizontal distance of the first bullet $=x_{A}$
The horizontal distance of the second bullet $=x_{B}$
The vertical distance of the first bullet $=h_{A}$
The vertical distance of the second bullet $=h_{B}$
Let's apply the equation $S=ut$ in the horizontal direction for both bullets.
Bullet 1 =>
$\rightarrow S=ut $; Let's plug known values into this equation.
$x_{A}=Vt_{A}-(1)$
Similarly for bullet 2 =>
$x_{B}=Vt_{B}-(2)$
Given that, $x_{B}=2x_{A}$
(2)=>
$2x_{A}=Vt_{B}-(3)$
Let's apply the equation $S=ut+\frac{1}{2}at^{2}$ in the vertical direction for both bullets.
Bullet 1 =>
$\downarrow S=ut +\frac{1}{2}at^{2}$; Let's plug known values into this equation.
$h_{A}=0+\frac{1}{2}gt_{A}^{2}=\frac{1}{2}gt_{A}^{2}-(4)$
Similarly,
$h_{B}=0+\frac{1}{2}gt_{B}^{2}=\frac{1}{2}gt_{B}^{2}-(5)$
(5)/(4) =>
$\frac{h_{B}}{h_{A}}=\frac{\frac{1}{2}gt_{B}^{2}}{\frac{1}{2}gt_{A}^{2}}= (\frac{t_{B}}{t_{A}})^{2}-(6)$
(1),(3)=>(6)
$\frac{h_{B}}{h_{A}}=(\frac{2x_{A}/V}{x_{A}/V})^{2}=4$
