Projectile from hill, height, speed, time

Take $g=9.81\ \text{m/s}^2$ and the origin at P, with $y$ upward.

Launch components

$$v_x=120\cos30^\circ=103.92\ \text{m/s},\qquad v_{y0}=120\sin30^\circ=60\ \text{m/s}$$

i) Maximum height above the level of projection

At the top $v_y=0$:

$$H=\frac{v_{y0}^{2}}{2g}=\frac{60^{2}}{2(9.81)}=183.49\ \text{m}$$

ii) Velocity on striking the target ($y=-100$ m)

$$v_y^{2}=v_{y0}^{2}+2g(100)=3600+1962=5562\ \Rightarrow\ v_y=74.58\ \text{m/s (downward)}$$ $$v=\sqrt{103.92^{2}+74.58^{2}}=127.9\ \text{m/s}$$

Direction: $\tan^{-1}(74.58/103.92)=35.7^\circ$ below the horizontal.

iii) Total time of flight

$$t=\frac{v_{y0}+|v_y|}{g}=\frac{60+74.58}{9.81}=13.72\ \text{s}$$

Answer: (i) $H=183.5$ m; (ii) $v=127.9$ m/s at $35.7^\circ$ below horizontal; (iii) $t=13.72$ s.