Throw a ball across a field and watch it. It doesn't fly straight and then suddenly drop — it traces a smooth arch, rising, slowing, curving over, and coming back down. The secret to understanding that arch is one idea from Galileo: think of the sideways motion and the up-down motion as two separate stories happening at the same time. Sideways, nothing pushes the ball, so it just drifts at a steady speed. Up and down, gravity is always tugging it back to Earth. Glue the two together and you get the curve.
Putting numbers on it
Two quantities set everything in motion: the launch speed $v_0$ (how hard you throw) and the launch angle $\theta$ (how steeply). The throw splits into a sideways part $v_0\cos\theta$ and an upward part $v_0\sin\theta$. The sideways part never changes. The upward part fades, stops at the top, then grows downward. On flat ground the ball lands at a distance called the range:
Range on level ground
$$R=\frac{v_0^2\sin 2\theta}{g}$$
Try a real number: throw at $v_0=20\text{ m/s}$ and $\theta=45^\circ$ on Earth ($g=9.81\text{ m/s}^2$). Since $\sin 90^\circ=1$, the ball travels $R=400/9.81\approx 41\text{ m}$ — about half a football field.
Going deeper
The full picture is two equations running together: horizontally $x(t)=v_0\cos\theta\,t$, and vertically $y(t)=h_0+v_0\sin\theta\,t-\tfrac12 gt^2$. The $\sin 2\theta$ in the range formula is why $45^\circ$ is the sweet spot on flat ground — it's the angle that best balances "throw it far sideways" against "keep it in the air long enough." In the sim, the Speed and Angle sliders are exactly $v_0$ and $\theta$; Gravity is $g$ (try the planet presets); Height is $h_0$; and Drag adds the air resistance that real throws feel.
Try this in the sim above
First, set the angle to $45^\circ$, then to $30^\circ$ and $60^\circ$ — notice the range is the same for those complementary pairs, but the $60^\circ$ shot flies higher and hangs longer. Next, open the Compare Angles mode and fire all five at once to see the $45^\circ$ arc reach farthest. Finally, switch gravity to Moon (1.62 m/s²) and fire: with weaker gravity the same throw sails roughly six times farther.
Section 03
Equations & Derivation
Governing Equations — Projectile Motion (no air resistance)
Decompose initial velocity. Horizontal: $v_{0x}=v_0\cos\theta$ (constant, no horizontal force). Vertical: $v_{0y}=v_0\sin\theta$ (decreasing due to gravity $g$).
2
Equations of motion. Integrate $a_x=0$ and $a_y=-g$ with initial conditions: $x=v_{0x}t$, $y=h_0+v_{0y}t-\frac{1}{2}gt^2$.
3
Maximum height. At peak, $v_y=0$: $t_{peak}=\dfrac{v_0\sin\theta}{g}$, so $H=h_0+\dfrac{v_0^2\sin^2\theta}{2g}$.
4
Time of flight. Set $y=0$ and solve quadratic: $T=\dfrac{v_0\sin\theta+\sqrt{v_0^2\sin^2\theta+2gh_0}}{g}$. For $h_0=0$: $T=\dfrac{2v_0\sin\theta}{g}$.
5
Range. $R=v_{0x}\cdot T=\dfrac{v_0^2\sin 2\theta}{g}$ (for $h_0=0$). Maximum range at $\theta=45°$. Complementary angles ($\theta$ and $90°-\theta$) give equal range.
6
Trajectory equation. Eliminate $t$: $y=h_0+x\tan\theta-\dfrac{gx^2}{2v_0^2\cos^2\theta}$ — a downward-opening parabola.
7
With air drag. Linear drag $\vec{F}_d=-k\vec{v}$ gives coupled ODEs: $\dot{v}_x=-kv_x/m$, $\dot{v}_y=-g-kv_y/m$. Solved numerically (RK4) in the simulation — range and height are reduced, optimal angle shifts below 45°.
Ref: Halliday, Resnick & Walker — Fundamentals of Physics, 10th Ed., Ch. 4 "Motion in Two and Three Dimensions"; Serway & Jewett, 8th Ed., §4.3; Kleppner & Kolenkow, 2nd Ed., §3.3.
Section 04
Frequently Asked Questions
A projectile (ball) launched with speed $v_0$ at angle $\theta$ from height $h_0$. Horizontal and vertical motions are independent — horizontal velocity is constant (no air resistance), vertical velocity changes due to gravity. The trajectory is a parabola. With drag enabled, the path becomes asymmetric. The "R vs θ" graph shows how range varies with launch angle — peaking at 45° with no drag.
Range $R=\frac{v_0^2\sin 2\theta}{g}$ is maximised when $\sin 2\theta=1$, i.e. $2\theta=90°$, so $\theta=45°$. At this angle the product of horizontal speed ($v_0\cos\theta$) and time of flight ($\propto\sin\theta$) is maximised. If the launch point is elevated above the landing point, the optimal angle shifts below 45°; try the "From Height" mode to see this.
Yes — for flat ground with no drag. Since $R\propto\sin 2\theta$, and $\sin 2\theta=\sin(180°-2\theta)$, angles $\theta$ and $90°-\theta$ produce identical ranges. For example, 30° and 60° both give the same $R$. The 60° shot reaches a higher peak and stays airborne longer; the 30° shot is flatter and faster.
Artillery and ballistics (historically the main application). Sports: basketball free throws, soccer kicks, javelin, golf, long jump. Water from a hose or fountain. Roller coaster launches. Spacecraft re-entry trajectories (where drag dominates). Even the path of a jumping frog or a thrown stone follows projectile kinematics.
Air drag (proportional to velocity) slows the projectile continuously, reducing both range and maximum height. The trajectory is no longer a symmetric parabola — the descent is steeper than the ascent. The optimal launch angle for maximum range drops below 45° (around 30–40° depending on drag). Enable the drag slider to compare with the ideal (no-drag) path shown as a dashed curve.
In a vacuum (no drag), mass cancels and all projectiles follow identical paths for the same $v_0$ and $\theta$. With air drag, a heavier ball (for the same size) has higher mass but similar drag force, so its deceleration $a=F_d/m$ is smaller — it travels farther. This is why shot puts travel farther than ping-pong balls at the same launch speed. The simulation uses linear drag; increase drag to explore this effect.
Resources: Khan Academy — Projectile Motion (khanacademy.org); HyperPhysics — Projectile Motion (hyperphysics.phy-astr.gsu.edu); MIT OCW 8.01 — Lecture 4.
Section 05
Common Misconceptions
❌ Horizontal and vertical motions affect each other during flight.
✅ They are completely independent. Horizontal motion is uniform (constant $v_x = v_0\cos\theta$); vertical motion is uniformly accelerated ($a_y = -g$). This independence is the core insight of projectile motion, first stated by Galileo. A bullet fired horizontally and one dropped from the same height hit the ground at the same time.
❌ At the highest point, the projectile has zero velocity.
✅ Only the vertical component $v_y = 0$ at the peak. The horizontal component $v_x = v_0\cos\theta$ is never zero (unless $\theta = 90°$). So the speed at the top is $|v| = v_0\cos\theta$, not zero. The projectile is still moving horizontally at maximum height.
📖 Serway & Jewett, 8th Ed., §4.3, Example 4.3.
❌ Maximum range always occurs at 45°, regardless of conditions.
✅ This is only true for level ground with no air drag. When the launch point is higher than the landing point, the optimal angle is less than 45°. With air resistance, the optimal angle is also below 45° (typically 30–40°). Use the "From Height" mode or enable drag to verify this.
❌ A projectile follows a straight line until it "runs out of speed" then falls straight down.
✅ This was Aristotle's incorrect view. In reality, the projectile follows a smooth parabolic arc from launch to landing — there is no point where it "stops" horizontally. Gravity acts continuously from the moment of launch, curving the path downward throughout the flight.
📖 Galileo — Discorsi (1638); Arons — A Guide to Introductory Physics Teaching, Ch. 5.
❌ A higher launch speed always means a longer flight time.
✅ Flight time depends on the vertical component of velocity: $T = 2v_0\sin\theta/g$. A very fast horizontal throw ($\theta\approx 0°$) has almost no vertical component and lands almost instantly. A slower throw at $\theta = 90°$ (straight up) has much longer airtime. Speed alone does not determine flight duration — the angle matters crucially.
📖 Halliday, Resnick & Walker, 10th Ed., §4-4.
Misconception research: Halloun & Hestenes (1985), Am. J. Phys. 53, 1043; Clement (1982), Am. J. Phys. 50, 66 — Students' preconceptions of projectile motion.