On the edge of a cliff

Let $t$ be the total time taken by the rock to reach the ground and $d$ be the horizontal distance. We have $-20=14\sin\theta t-\frac{g}{2}t^2\\t=\frac{14\sin\theta+\sqrt{196\sin^2\theta+40g}}{g}\\d=t\cdot14\cos\theta=\frac{14\sin\theta+\sqrt{196\sin^2\theta+40g}}{g}\cdot14\cos\theta$

Differentiating with respect to $\theta$ , we have $\large\frac{\text{d}d}{\text{d}\theta}=\frac{\frac{2744\sin\theta\cos^2\theta}{\sqrt{196\sin^2\theta+40g}}-14\sin\theta\sqrt{196\sin^2\theta+40g}+196(\cos^2\theta-196\sin\theta)^2}{g}$

Setting it to zero, we find that $\theta\approx 30^\circ$ .

Clarification: The final messy equation was solved using wolfram alpha.

The angle for maximum range when thrown from a certain height can be put as

tan ^-1 (1/sqrt(1+(2H/L))) where H is the height from the horizontal plane (20 m in this case) and L = u^2/g (u is 14 m/s in this case)

Solving the above, we get the angle to be 30 degrees. :)

just differentiate trajectory equation with respect to theta ,,, and put dR/dtheta = 0 !!!!!!!