Let $\theta$ be the angle of elevation
$x(t) = x_0 + u_0t + \frac{1}{2} at^2$
where $x(t)=0$ and $x_0=50m$ , And $u_0$ is vertically resolved initial component of the velocity
I applied this for the vertical resolved motion which gives:
$-50 = (-600\sin \theta) t - \frac{1}{2}gt^2$
Then for the horizontal (horizontal component) motion I used :
$x = vt$
Which gives:
$12000 = (600 \cos \theta) t$
$t=\frac{1200}{600 \cos \theta}$
Substitute this as $t$ for the first equation:
$$-50=(- 600\sin \theta) \frac{1200}{ 600 \cos \theta}-5 (\frac{1200}{600 \cos \theta})^2$$
$$-50=(- \sin \theta) \frac{1200}{ \cos \theta}- \frac{20}{ \cos^2 \theta}$$
Multiplying each term by $cos^2 \theta$
$$-50 \cos^2 \theta=- 1200 \sin \theta \cos \theta- 20$$
How to solve this equation, Please help. Thanks.
$-50 \cos^2 \theta=- 1200 \sin \theta \cos \theta- 20$
Divide by $\cos^2\theta $ :
$-50 =- 1200 \tan \theta- 20(1+\tan^2\theta)$
Notice that its just a quadratic in $\tan \theta$, you can solve it by using quadratic formula