Answer:

When the x and y axes are set in the manner depicted in the diagram, the missile moves from point O to point A.

y = 0

u_y = v_o cos θ

a_y = -g cos θ

t = T

Using the kinematics for the y-axis

y = u_yt + 1/2 a_yt²

T = 2v_o cos θ/g cos θ

Using kinematics for the x-axis

x = u_xt + 1/2 a_xt²

Solving for the distance,

$L =\left[\frac{2 v _{0}}{ g }\right] v _{0} \sin \theta+\frac{1}{2} g \sin \theta\left[\frac{2 v _{0}}{ g }\right]^{2}$
$L =\frac{2 v _{0}^{2}}{ g } \sin \theta+\frac{1}{2} g \sin \theta \cdot \frac{4 v _{0}^{2}}{ g ^{2}}$
$=\frac{2 v _{0}^{2}}{ g }[\sin \theta+\sin \theta]=\frac{2 v _{0}^{2}}{ g } 2 \sin \theta$
$\Longrightarrow L =\frac{4 v _{0}^{2}}{ g } \sin \theta$