A projectile moves under gravity alone (no air resistance), giving constant horizontal velocity and constant downward acceleration $g$.
Launch from origin at speed $V$, angle $\alpha$ above horizontal:
$$\mathbf{u} = V\cos\alpha\,\mathbf{i} + V\sin\alpha\,\mathbf{j}$$
$$\mathbf{a} = -g\mathbf{j} \quad (\text{taking up as positive})$$
Equations of motion:
$$x(t) = V\cos\alpha\cdot t$$
$$y(t) = V\sin\alpha\cdot t - \frac{1}{2}gt^2$$
Velocity:
$$\dot x = V\cos\alpha, \qquad \dot y = V\sin\alpha - gt$$
Eliminate $t$: from $x = V\cos\alpha\cdot t$, $t = \dfrac{x}{V\cos\alpha}$. Substitute:
$$y = x\tan\alpha - \frac{gx^2}{2V^2\cos^2\alpha}$$
This is a downward-opening parabola in the $xy$-plane.
Time of flight ($y = 0$, $t > 0$):
$$t_f = \frac{2V\sin\alpha}{g}$$
Range (horizontal distance when $y = 0$):
$$R = V\cos\alpha \cdot t_f = \frac{V^2\sin 2\alpha}{g}$$
Range is maximised when $\sin 2\alpha = 1 \Rightarrow \alpha = 45^\circ$.
Maximum height (when $\dot y = 0$):
$$t_{\text{max}} = \frac{V\sin\alpha}{g}, \qquad H = \frac{V^2\sin^2\alpha}{2g}$$
Speed at any time: $|\mathbf{v}| = \sqrt{(V\cos\alpha)^2 + (V\sin\alpha - gt)^2}$.
At maximum height: $|\mathbf{v}| = V\cos\alpha$ (horizontal component only).
A ball is launched at $15$ m/s at $30^\circ$ above horizontal from a cliff 20 m high. Find (a) the time it hits the ground, (b) horizontal range from the base of the cliff.
Let ground be $y = -20$ (cliff top at origin). $g = 9.8$ m/s$^2$.
$u_x = 15\cos30^\circ = \dfrac{15\sqrt{3}}{2}$, $u_y = 15\sin30^\circ = 7.5$ m/s.
(a) $-20 = 7.5t - 4.9t^2 \Rightarrow 4.9t^2 - 7.5t - 20 = 0$.
$$t = \frac{7.5 + \sqrt{56.25 + 392}}{9.8} = \frac{7.5 + \sqrt{448.25}}{9.8} \approx \frac{7.5+21.17}{9.8} \approx 2.93 \text{ s}$$
(b) $x = \dfrac{15\sqrt{3}}{2} \times 2.93 \approx 12.99 \times 2.93 \approx 38.1$ m.
Using $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$:
$$\mathbf{r}(t) = (V\cos\alpha\, t)\mathbf{i} + \left(V\sin\alpha\, t - \tfrac{g}{2}t^2\right)\mathbf{j}$$
This compact form handles launch from any initial position by replacing $\mathbf{r}_0 = \mathbf{0}$ with the actual launch point.
KEY TAKEAWAY: Projectile motion decomposes into independent horizontal (constant velocity) and vertical (constant acceleration) components. The trajectory is always a parabola.
EXAM TIP: When the landing point is not at the same height as launch, set up the $y$-equation and solve for $t$ first, then compute $x$. Do not use range formula in this case.
COMMON MISTAKE: Using the range formula $R = V^2\sin2\alpha/g$ when launch and landing heights differ — the formula only applies to level ground.
