Two projectiles at an initial height with a missing angle of projection 
David and Goliath are testing what they learned in school, that the
  distance travelled by launching a projectile is dependent on a range
  of variables; the launch angle, height and velocity of projection. 
As Goliath is significantly taller, David decides to throw a small
  projectile while standing on a ladder so that he is throwing from the
  same vertical height and same spot as Goliath.
Goliath launches first with an angle of 60$^{\circ}$ to the
  horizontal. David will then launch the projectile next with the same
  velocity of projection as Goliath. 
If David wants to match Goliath’s attempt and achieve the same
  horizontal distance, at what angle should he launch the projectile if
  he can’t do it at an angle of 60$^{\circ}$?
(Use $ = −10ms^{-1}$ in your calculations and use this as the only
  force acting upon the projectile.)

If David's projectile's displacement is given by $r_d =  v t \cos \theta \hat{i} + (h + v t \sin \theta - 5t^2) \hat{j}$ then David's projectile hits the ground when $\frac{v \sin \theta \pm \sqrt{v^2 \sin^2\theta+20h}}{10} = 0$. Solving this, the time it takes David's projectile to reach the ground is $T_d = \frac{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}{10}$.
If Goliath's projectile's displacement is given by $r =  \tfrac{v}{2} t \hat{i} + (h + \tfrac{\sqrt{3}v}{2}t - 5t^2) \hat{j}$ then Goliath's projectile hits the ground when $\frac{\sqrt{3}v \pm \sqrt{3v^2 + 80h}}{20} = 0$. Solving this, the time it takes Goliath's projectile to reach the ground is $T_g = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}$.
For the horizontal distance travelled to match, their terminal horizontal components must match. Hence, 
\begin{align*}
    v T_d \cos \theta   &= \tfrac{v}{2} T_g\\
    T_d \cos \theta     &= \tfrac{1}{2} T_g\\
    \cos \theta     &= \frac{T_g}{2} \frac{1}{T_d}\\
    &= \frac{\frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{20}}{2} \frac{10}{v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}}\\
   \cos \theta &= \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) }\\
\end{align*}
I'm having a major mental block on how to obtain an expression for $\theta$ in terms of $h$ and $v$. For what its worth, wolframalpha gives $\theta = -\cos^{-1}\left(-\frac{\sqrt{200 h^2 + 70 h v^2 + 10 \sqrt{3} h v \sqrt{80 h + 3 v^2} + \sqrt{3} v^3 \sqrt{80 h + 3 v^2} + 3 v^4}}{\sqrt{800 h^2 + 80 h v^2 + 8 v^4}}\right)$. I seem to missing out some details in my calculations.
What's the simplest way to obtain the formula for $\theta$ in terms of $v$ and $h$?
 A: From
\begin{equation*}
\cos \theta = \frac{\sqrt{3}v + \sqrt{3v^2 + 80h}}{4 (v \sin \theta + \sqrt{v^2 \sin^2\theta+20h}) },
\end{equation*}
multiplying both numerator and denominator by $v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h}$ gives
\begin{align*}
\cos \theta &= \frac{(\sqrt{3}v + \sqrt{3v^2 + 80h})(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta} + 20h})}{4 (-20h) }\\
&= A(v\sin{\theta} - \sqrt{v^{2}\sin^{2}{\theta}+20h}),
\end{align*}
where we have set $A= -\dfrac{\sqrt{3}v + \sqrt{3v^{2} + 80h}}{80h}.$ We then get
\begin{align*}
\Big(\frac{\cos{\theta}}{A} - v\sin{\theta}\Big)^{2} &= v^{2}\sin^{2}{\theta} + 20h,
\end{align*}
or equivalently
\begin{align*}
\frac{\cos^{2}{\theta}}{A^{2}} - 2\frac{v\cos{\theta}\sin{\theta}}{A} = 20h.
\end{align*}
Thus
\begin{equation*}
(\cos^{2}{\theta} - 20A^{2}h)^{2} = 4A^{2}v^{2}\cos^{2}{\theta}(1-\cos^{2}{\theta}).
\end{equation*}
Expanding this equation gives the quadratic equation in $\cos^{2}{\theta}$
\begin{equation*}
(1+4A^{2}v^{2})\cos^{4}{\theta}- 4A^{2}(10h + v^{2})\cos^{2}{\theta} + 400 A^{4}h^{2} = 0,
\end{equation*}
and solving for $\cos^{2}{\theta}$ yields
\begin{align*}
\cos^{2}{\theta} &= \frac{4A^{2}(10h+v^{2})\pm\sqrt{16A^{4}(10h+v^{2})^{2}-4(400A^{4}h^{2})(1+4A^{2}v^{2})}}{2(1+4A^{2}v^{2})} \\[1ex]
&= \frac{2A^{2}(10h+v^{2})\pm 2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}.
\end{align*}
There will be two solutions for $\theta$, one of them being at $\theta = 60^{\circ}$ which we aren't interested in, and the other being at some angle less than $60^{\circ}$. To get the smaller angle, we take the positive sign in the above expression since $\cos^{2}{\theta}$ is decreasing on $\Big(0, \dfrac{\pi}{2}\Big)$. Finally, as $\cos{\theta} > 0$, we obtain
\begin{align*}
\cos{\theta} = \sqrt{\frac{2A^{2}(10h+v^{2})+2A^{2}\sqrt{(10h+v^{2})^{2}-100h^{2}(1+4A^{2}v^{2})}}{1+4A^{2}v^{2}}}.
\end{align*}
A: I got a very different answer.
I start with the following formula given here:
https://en.wikipedia.org/wiki/Range_of_a_projectile
$d = \frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right)$
Now we must have
$\frac{v^2 \sin 2 \theta}{2g}\left(1 + \sqrt{1+ \frac{2gy_0}{v^2 \sin^2 \theta}}\right) = \frac{v^2 \sin 120^\circ}{g} = \frac{\sqrt{3}v^2}{2g}$
Solving,
$\theta = \frac{1}{2} \sin^{-1} \frac{3v}{\sqrt{6v^2 + 20y_0}}$
Please check the calculations.
