Projectiles

$u_x = 20\cos 30° = 17.3$ and $u_y = 20\sin 30° = 10\ \text{m s}^{-1}$.

$T = \dfrac{2u\sin\alpha}{g} = \dfrac{2(10)}{9.81} = 2.04$ s.

$R = \dfrac{u^2\sin 2\alpha}{g} = \dfrac{400\sin 60°}{9.81} = 35.3$ m and $H = \dfrac{u_y^2}{2g} = \dfrac{100}{19.62} = 5.10$ m.

$R = \dfrac{u^2\sin 2\alpha}{g}$ is greatest when $\sin 2\alpha = 1$, that is $2\alpha = 90°$ and $\alpha = 45°$.

$45 = \tfrac12(9.81)t^2$ gives $t^2 = 9.174$ and $t = 3.03$ s. Then $x = 12(3.03) = 36.3$ m.

$R$ depends on $\sin 2\alpha$. For $30°$, $2\alpha = 60°$; for $60°$, $2\alpha = 120°$. Since $\sin 60° = \sin 120°$, the ranges are equal — though the $60°$ path is higher and slower.