Answer
$27.2^{\circ}$
Work Step by Step
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time of projectile 1. (Initial velocity $V$, launch angle to horizontal $\theta_{1}$)
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$0=Vsin\theta_{1}t+\frac{1}{2}(-g)t^{2}$
$t=\frac{2Vsin\theta_{1}}{g}$
Let's apply the equation $S=ut$ in the horizontal direction to find the range of projectile 1.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$R_{2}=Vcos\theta_{1}\times\frac{2Vsin\theta_{1}}{g}=\frac{V^{2}sin2\theta_{1}}{g}-(1)$
Similarly,
$R_{2}=\frac{V^{2}sin2\theta_{2}}{g}-(2)$
(2)/(1)=>
$\frac{R_{2}}{R_{1}}=\frac{\frac{V^{2}sin2\theta_{2}}{g}}{\frac{V^{2}sin2\theta_{1}}{g}}=2$
$\frac{sin2\theta_{2}}{sin2\theta_{1}}=2$ ;Let's plug $\theta_{1}=12^{\circ}$
$sin2\theta_{2}=2\times sin24^{\circ}=0.813$
$\theta_{2}=\frac{sin^{-1}(0.813)}{2}=27.2^{\circ}$
