Projectile Motion
A body thrown into space moving under gravity alone traces a parabola. Explore every concept with live diagrams, full derivations and a hands-on launch simulator — then test yourself with real exam questions.
Motion in a Plane
When a body moves such that its instantaneous position can be described in two mutually perpendicular directions, the motion is said to be two-dimensional (motion in a plane) — for example a projectile or circular motion.
The angle between the initial velocity and the acceleration decides the shape of the trajectory:
| Angle between velocity & acceleration | Path of motion |
|---|---|
| $0^\circ$ or $180^\circ$ | Straight line (rectilinear) |
| $90^\circ$ | Circular |
| Any other angle | Parabolic (curvilinear) |
- The velocity of a projectile changes in both magnitude and direction.
- Its motion is the superposition of a uniform horizontal motion and a uniformly accelerated vertical motion.
- Acceleration is constant ($=g$, vertically down) throughout the flight.
Horizontal Projection from a Height
A particle is projected horizontally with speed $u$ from a height $H$. The horizontal and vertical motions are independent.
$$x = ut, \qquad y = H - \tfrac{1}{2}gt^2$$ $$\text{Trajectory: } y = H - \dfrac{g}{2u^2}\,x^2 \quad(\text{a parabola})$$Full derivation — trajectory, time of flight & range
Take the launch point at the top of a cliff of height $H$, the $x$-axis horizontal and $y$ measured upward from the ground. The initial velocity is purely horizontal, $u_x=u,\ u_y=0$, and the only acceleration is $g$ downward.
Step 1 — Independent motions
Horizontally there is no force, so the motion is uniform; vertically the body falls freely from rest:
$$x = ut, \qquad H - y = \tfrac{1}{2}gt^2 \;\Rightarrow\; y = H - \tfrac{1}{2}gt^2$$Step 2 — Equation of the path
Eliminate $t$ using $t = x/u$:
$$y = H - \tfrac{1}{2}g\left(\frac{x}{u}\right)^2 = H - \frac{g}{2u^2}x^2$$$y$ is quadratic in $x$, so the path is a parabola.
Step 3 — Time of flight
The body lands when $y=0$, i.e. it has fallen the full height $H$:
$$H = \tfrac{1}{2}gT^2 \;\;\Rightarrow\;\; T = \sqrt{\frac{2H}{g}}$$Step 4 — Horizontal range
The constant horizontal speed carried for time $T$ gives:
Step 5 — Velocity on landing
$$v = \sqrt{v_x^2+v_y^2} = \sqrt{u^2+g^2t^2}, \qquad \alpha = \tan^{-1}\!\frac{v_y}{v_x} = \tan^{-1}\!\frac{gt}{u}$$Drop vs Launch — who lands first?
A ball dropped straight down and a ball launched horizontally leave the same height at the same instant. Change the launch speed — they still touch the ground together, because the vertical fall is independent of the horizontal motion.
Projection at an Angle
A body is projected from the ground at angle $\theta$ to the horizontal with speed $u$. Resolve into $u_x=u\cos\theta$ (constant) and $u_y=u\sin\theta$ (decelerated by $g$).
$$x = u\cos\theta\,t, \qquad y = u\sin\theta\,t - \tfrac{1}{2}gt^2$$ $$\text{Trajectory: } y = x\tan\theta - \dfrac{g}{2u^2\cos^2\theta}\,x^2$$Full derivation — time of flight, max height & range
Project a body from the origin at angle $\theta$ with speed $u$. The horizontal component $u\cos\theta$ is unchanged; the vertical component $u\sin\theta$ is retarded by gravity $g$.
Step 1 — Time to reach the highest point
At the top the vertical velocity is zero:
$$0 = u\sin\theta - g\,t' \;\;\Rightarrow\;\; t' = \frac{u\sin\theta}{g}$$Step 2 — Total time of flight
By symmetry the up and down journeys take equal time, so:
$$T = 2t' = \frac{2u\sin\theta}{g}$$Step 3 — Maximum height
Using $v^2=u^2-2gH$ on the vertical motion with final vertical velocity $0$:
$$H = \frac{(u\sin\theta)^2}{2g} = \frac{u^2\sin^2\theta}{2g}$$Step 4 — Horizontal range
The horizontal speed acts for the whole flight time $T$:
$$R = u\cos\theta \times T = u\cos\theta\cdot\frac{2u\sin\theta}{g} = \frac{u^2(2\sin\theta\cos\theta)}{g}$$Step 5 — Equation of the trajectory
Eliminating $t$ from $x=u\cos\theta\,t$ and $y=u\sin\theta\,t-\tfrac12 gt^2$:
$$y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}x^2$$Quadratic in $x$ — confirming the path is a parabola.
Resolve the launch velocity
Drag the angle and speed to watch the launch velocity $u$ split into a constant horizontal part $u\cos\theta$ and a vertical part $u\sin\theta$ that gravity steadily slows.
Interactive Launch Simulator
Maximum Height & Maximum Range
- Range is maximum when $\sin2\theta=1$, i.e. at $\theta=45^\circ$: $$R_{max} = \dfrac{u^2}{g}$$
- Maximum possible height (thrown straight up, $\theta=90^\circ$): $$H_{max} = \dfrac{u^2}{2g} = \dfrac{R_{max}}{2}$$
- Complementary angles $\theta$ and $(90^\circ-\theta)$ give the same range with the same speed.
Derivation — angle for maximum range & complementary angles
The range of a ground-to-ground projectile is $R=\dfrac{u^2\sin2\theta}{g}$ for a fixed launch speed $u$.
Step 1 — Maximise the range
For fixed $u$, $R$ is largest when $\sin2\theta$ is largest. Since $\sin2\theta\le1$:
$$\sin2\theta = 1 \;\Rightarrow\; 2\theta = 90^\circ \;\Rightarrow\; \theta = 45^\circ$$Step 2 — Two angles, one range
Replace $\theta$ by its complement $(90^\circ-\theta)$:
$$R(90^\circ-\theta) = \frac{u^2\sin\big(2(90^\circ-\theta)\big)}{g} = \frac{u^2\sin(180^\circ-2\theta)}{g} = \frac{u^2\sin2\theta}{g} = R(\theta)$$So, e.g., $15^\circ$ and $75^\circ$ — or $30^\circ$ and $60^\circ$ — share the same range, while the steeper angle gives the greater height and longer flight time.
Range vs angle — find the maximum
The range follows $R=\dfrac{u^2}{g}\sin2\theta$. Slide the angle to trace the curve: the peak sits exactly at $45^\circ$, and every angle shares its range with its complement $(90^\circ-\theta)$.
Energy & Quantities Along the Path
A projectile of mass $m$ is thrown at speed $u$ and angle $\theta$ with initial kinetic energy $E_0=\tfrac12 mu^2$. At the highest point only the horizontal velocity survives:
- Speed, velocity, momentum & KE first decrease to a minimum at the top, then increase again.
- Potential energy increases to a maximum at the top, then decreases.
- Mechanical energy stays constant and acceleration ($=g$) is constant throughout.
- The angle between velocity and acceleration decreases gradually from $(90^\circ+\theta)$ to $(90^\circ-\theta)$; at the top velocity ⟂ acceleration.
Launch ⟶ Highest point (O to C)
- Change in speed $= u(1-\cos\theta)$
- Change in velocity $= u\sin\theta$
- Change in momentum $= mu\sin\theta$
- Decrease in KE $=\tfrac12 mu^2\sin^2\theta = E_0\sin^2\theta$
- Increase in PE $=\tfrac12 mu^2\sin^2\theta = E_0\sin^2\theta$
- Change in direction $= \theta$
Launch ⟶ Landing (O to B)
- Change in speed $= 0$
- Change in velocity $= 2u\sin\theta$
- Change in momentum $= 2mu\sin\theta$
- Change in KE $= 0$
- Change in PE $= 0$
- Change in direction $= 2\theta$
Projectile on an Inclined Plane
A projectile is thrown with speed $u$ at angle $\alpha$ to the horizontal, up an incline of inclination $\beta$ (with $\beta<\alpha$). Resolve along and perpendicular to the incline.
Effect of Air Resistance
When air resistance is not neglected, the symmetric parabola is broken:
- Maximum height, horizontal range, striking speed, momentum and KE are each less than in the ideal (drag-free) case.
- The time of flight is longer than without air resistance.
- The angle at which the projectile strikes the ground is greater than the angle of projection — the descent is steeper than the ascent.