You met sideways throws in Horizontal Launches and angled ones in Launches at an Angle. An exam problem strings the pieces together: first the time in the air, then the distance, then the speed or direction at landing. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Split the motion. Sideways, no force: $x = (v_0\cos\theta)\,t$. Up and down, acceleration $g$: $y = (v_0\sin\theta)\,t - \tfrac{1}{2}gt^2$. The two share only the time $t$. For the landing speed, energy is quickest: $$v = \sqrt{v_0^2 + 2gh},$$ whatever the launch angle.
Worked example: A stone from a cliff
A stone is thrown horizontally at $15\ \text{m/s}$ from a cliff $45\ \text{m}$ above the sea. Find (1) the time in the air, (2) how far out it lands and (3) its speed at the water.
- (1) Vertically it starts from rest: $45 = \tfrac{1}{2}(9.8)t^2$, so $t = \sqrt{90/9.8} = 3.03\ \text{s}$.
- (2) Sideways the speed stays $15\ \text{m/s}$: $x = 15(3.03) = 45.5\ \text{m}$.
- (3) Energy: $v = \sqrt{15^2 + 2(9.8)(45)} = \sqrt{1107} = 33.3\ \text{m/s}$.
Answer: $3.03\ \text{s}$, $45.5\ \text{m}$, $33.3\ \text{m/s}$
Common mistake
Using the whole launch speed in the vertical equation for an angled throw. Only $v_0\sin\theta$ is vertical. Check: a throw straight sideways ($\theta = 0$) must take the same time to fall as a dropped stone.