A ball launched toward a wall, a net or a fence asks two things at once: when does it get there, and where is it then? The sideways motion answers the first, the vertical motion the second. Builds on Launches at an Angle. 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
$$t = \frac{d}{v_0\cos\theta}, \qquad y = (v_0\sin\theta)t - \tfrac{1}{2}gt^2, \qquad v_y = v_0\sin\theta - gt.$$ The sign of $v_y$ says whether the ball is still rising ($+$) or already falling ($-$). The top of the flight is at $H_{\max} = \dfrac{(v_0\sin\theta)^2}{2g}$.
Worked example: Over the wall
A ball is kicked at $20\ \text{m/s}$, $40^\circ$ above level ground, toward a wall $18\ \text{m}$ away. Find (1) when it reaches the wall, (2) its height there and (3) whether it is rising or falling.
- (1) $v_x = 20\cos 40^\circ = 15.3\ \text{m/s}$, so $t = 18/15.3 = 1.17\ \text{s}$.
- (2) $v_0\sin\theta = 12.9\ \text{m/s}$; $y = 12.9(1.17) - 4.9(1.17)^2 = 8.34\ \text{m}$.
- (3) $v_y = 12.9 - 9.8(1.17) = 1.34\ \text{m/s}$: positive, still rising.
Answer: $1.17\ \text{s}$, $8.34\ \text{m}$, rising
Common mistake
Assuming the ball must still be rising at the wall, or at its peak there. Work out $v_y$ and look at its sign; the peak is at $t = v_0\sin\theta/g$, which may come before or after the wall.