Answer
$41.5^{\circ}$
Work Step by Step
By using trigonometry, we can get.
$tan\theta=(\frac{2400\space m}{x})-(1)$ ; Where x is the horizontal displacement of the flare.
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$-2400=-240sin30^{\circ}t+\frac{1}{2}(-9.8)t^{2}$
$4.9t^{2}+120t-2400=0$ ; This is a quadratic formula. So we can obtain the solutions as follows.
$t=\frac{-120\space\pm\sqrt {(-120)^{2}-4(4.9)(-2400)}}{2(4.9)}=\frac{-120\space\pm247.9}{y}$
Here we neglect the negative solution & we get $t=13.05\space s$
Let's apply the equation $S=ut$ in the horizontal direction to find the x
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$x=240\space m/s\times cos30^{\circ}\times13.05=2712.4\space m$
(1)=>
$\theta=tan^{-1}(\frac{2400\space m}{2712.4\space m})=41.5^{\circ}$
