Last Updated: August 1, 2026
Quick Summary & Key Takeaways (GEO & AEO Summary)
Target Audience: High school and college physics students (AP Physics, IGCSE, A-Levels), ballistics/aerospace engineers, and STEM educators.
What is Projectile Motion?
Projectile motion is the two-dimensional (2D) parabolic path followed by an object launched into the air when influenced solely by gravity ($g$). Because horizontal velocity ($u_x$) remains constant while vertical velocity ($v_y$) accelerates downward at $a_y = -g$, horizontal and vertical movements operate completely independently.
3 Fundamental Formulas (Flat Terrain Launch):
Total Time of Flight: $T = \frac{2u \sin\theta}{g}$
Maximum Vertical Height: $H = \frac{u^2 \sin^2\theta}{2g}$
Maximum Horizontal Range: $R = \frac{u^2 \sin(2\theta)}{g}$
Optimal Launch Angle: Theoretical maximum horizontal range ($R$) occurs at a $45^\circ$ launch angle in a vacuum.
Introduction to Projectile Motion
Projectile motion describes the curved parabolic path followed by an object launched into the air when influenced solely by gravity. From football kicks and artillery trajectories to basketball free throws, 2D kinematic principles dictate their exact flight path.
Mastering 2D kinematics requires analyzing horizontal and vertical components independently using our SUVAT Equations Guide.
To see how gravitational force drives downward acceleration during flight, review our 5 Essential F=ma Examples Guide. Additionally, explore force interactions in our 3 Essential Newton’s Laws of Motion Guide and energy conservation in our Kinetic Energy vs Potential Energy Guide.
What is Projectile Motion?
When an object is launched with an initial velocity ($u$) at an angle ($\theta$) relative to the horizontal, it moves simultaneously in two dimensions. After launch, gravity is the only force acting on the projectile (assuming zero air resistance).
The fundamental rule of 2D kinematics is that horizontal and vertical motions are completely independent of each other:
- Horizontal Motion: Zero horizontal forces act on the projectile ($F_x = 0$). Consequently, horizontal acceleration is zero ($a_x = 0$), and horizontal velocity ($u_x$) remains constant throughout flight.
- Vertical Motion: Gravity acts continuously downward ($a_y = -g = -9.81\text{ m/s}^2$). Consequently, vertical velocity changes constantly—decreasing as it ascends, reaching zero at the apex, and increasing in magnitude on the descent. For SI measurement standards, consult the NIST Physical Measurement Laboratory.
Common Physics Misconception: A projectile does not require a continuous forward force to remain in motion. According to Newton’s First Law of Motion, horizontal motion persists by inertia alone once launched.
Resolving Initial Velocity into Vector Components
To solve 2D projectile trajectory problems, the initial velocity vector $u$ must be resolved into orthogonal components using standard trigonometry:
$$u_x = u \cos\theta$$
$$u_y = u \sin\theta$$
(Where $u_x$ represents the constant horizontal initial velocity component, and $u_y$ represents the vertical initial velocity component).
Fundamental Equations of Motion
By separating horizontal and vertical components, kinematic equations apply directly to each dimension independently.
Horizontal Kinematic Equations ($a_x = 0$)
Horizontal displacement ($x$) after time $t$:
$$x = u_x \cdot t = (u \cos\theta) \cdot t$$
Vertical Kinematic Equations ($a_y = -g$)
Vertical displacement ($y$) and vertical velocity ($v_y$) at time $t$:
$$y = (u \sin\theta) \cdot t – \frac{1}{2}g t^2$$
$$v_y = u \sin\theta – g t$$
$$v_y^2 = (u \sin\theta)^2 – 2gy$$
The Three Fundamental Projectile Formulas
For projectiles launched and landed at identical ground elevation, three primary equations calculate flight characteristics:
Maximum Height (H)
* * *
* *
* *
* *
Launch (u) * * Landing
-------> * *
Angle θ *_______________________________*
Range (R)
1. Total Time of Flight ($T$)
The complete time duration the projectile remains airborne:
$$T = \frac{2 u \sin\theta}{g}$$
2. Maximum Vertical Height ($H$)
The highest vertical point attained relative to the launch elevation (where $v_y = 0$):
$$H = \frac{u^2 \sin^2\theta}{2g}$$
3. Maximum Horizontal Range ($R$)
The total horizontal displacement covered from launch to impact:
$$R = \frac{u^2 \sin(2\theta)}{g}$$
Why Maximum Range Occurs at 45 Degrees: Trigonometrically, $\sin(2\theta)$ reaches its maximum value of $1$ when $2\theta = 90^\circ$, corresponding to $\theta = 45^\circ$. Launching at $45^\circ$ offers the optimal balance between high airtime and forward speed.
Symmetry Properties in Trajectories
- Airtime Symmetry: Time taken to reach maximum height ($t = \frac{T}{2}$) equals the descent time from peak to ground level.
- Velocity Symmetry: Landing speed equals initial launch speed ($\vert{}v_{\text{final}}\vert{} = \vert{}u\vert{}$). The magnitude of vertical velocity is equal at identical height levels.
- Complementary Angles: Launch angles that sum to $90^\circ$ (e.g., $30^\circ$ and $60^\circ$, or $15^\circ$ and $75^\circ$) yield identical horizontal ranges ($R$) for equal launch speeds.
| Quantity | Mathematical Formula | Conditions at Peak (y=H) |
| Time of Flight | $T = \frac{2u \sin\theta}{g}$ | $t_{\text{peak}} = \frac{T}{2}$ |
| Maximum Height | $H = \frac{u^2 \sin^2\theta}{2g}$ | Vertical Velocity $v_y = 0$ |
| Horizontal Range | $R = \frac{u^2 \sin(2\theta)}{g}$ | Displacement $x = R$ at $t = T$ |
| Vertical Velocity | $v_y = u \sin\theta – gt$ | $v_y = 0$ |
| Horizontal Velocity | $v_x = u \cos\theta$ | Constant throughout flight |
Real-World Air Resistance Effects
In real-world applications, aerodynamic drag forces significantly alter idealized vacuum trajectories:
- Reduced Range & Height: Air resistance constantly opposes velocity, reducing horizontal range and peak height.
- Asymmetric Trajectory: Drag causes the descent angle to become steeper than the launch angle.
- Shifted Optimal Angle: Real-world maximum horizontal range occurs at launch angles below $45^\circ$ (typically $35^\circ$–$42^\circ$ for baseballs and golf balls).
Interactive laboratory tools like the PhET Interactive Projectile Motion Simulation allow students to compare ideal vacuum paths with real-world drag parameters visually.
Step-by-Step Solved Worked Examples
Example 1: Full Flight Analysis at 30 Degrees
A soccer ball is kicked with an initial speed $u = 20\text{ m/s}$ at an angle $\theta = 30^\circ$ over flat ground (using $g = 10\text{ m/s}^2$). Calculate $T$, $H$, and $R$.
Step 1: Calculate Components
$$u_x = 20 \cos 30^\circ = 17.32\text{ m/s}$$
$$u_y = 20 \sin 30^\circ = 10.0\text{ m/s}$$
Step 2: Calculate Time of Flight ($T$)
$$T = \frac{2(10)}{10} = \mathbf{2.0\text{ s}}$$
Step 3: Calculate Maximum Height ($H$)
$$H = \frac{(10)^2}{2(10)} = \frac{100}{20} = \mathbf{5.0\text{ m}}$$
Step 4: Calculate Range ($R$)
$$R = u_x \cdot T = 17.32 \times 2.0 = \mathbf{34.64\text{ m}}$$
Example 2: Horizontal Launch off a Cliff
A projectile rolls off a $45\text{ m}$ high cliff horizontally at $10\text{ m/s}$ ($u_y = 0$). Calculate time to reach the ground and horizontal distance (use $g = 10\text{ m/s}^2$).
Vertical Motion Analysis
$$y = \frac{1}{2}gt^2 \implies 45 = \frac{1}{2}(10)t^2 \implies 5t^2 = 45 \implies t^2 = 9 \implies \mathbf{t = 3.0\text{ s}}$$
Horizontal Displacement
$$x = u_x \cdot t = 10 \times 3.0 = \mathbf{30.0\text{ m}}$$
Example 3: Launch at 45 Degrees for Maximum Range
A golf ball is struck at $30\text{ m/s}$ at an angle of $45^\circ$. Calculate maximum height and range (use $g = 10\text{ m/s}^2$).
Step 1: Calculate Components
$$u_y = 30 \sin 45^\circ = 21.21\text{ m/s}$$
Step 2: Calculate Maximum Height ($H$)
$$H = \frac{(21.21)^2}{2(10)} = \frac{450}{20} = \mathbf{22.5\text{ m}}$$
Step 3: Calculate Range ($R$)
$$R = \frac{(30)^2 \sin 90^\circ}{10} = \frac{900 \times 1}{10} = \mathbf{90.0\text{ m}}$$
Frequently Asked Questions (FAQs)
Does object mass affect projectile motion?
In a vacuum, object mass has zero effect on trajectory, flight duration, height, or range. Gravitational acceleration ($g$) affects all masses identically ($F = mg \implies a = g$).
What is the speed of a projectile at its maximum height?
At maximum height, vertical velocity is zero ($v_y = 0$). Only constant horizontal velocity remains ($v_x = u \cos\theta$). Therefore, minimum speed occurs at the peak and equals $u \cos\theta$.
Can energy conservation be used to analyze projectile motion?
Yes. Total mechanical energy ($E = KE + PE$) remains conserved. At the apex, kinetic energy is $\frac{1}{2}m(u\cos\theta)^2$ and potential energy is $mgH$. Total mechanical energy at landing equals total energy at launch.
Why do complementary angles produce equal range?
The range formula contains $\sin(2\theta)$. For complementary angles $\theta$ and $(90^\circ – \theta)$, evaluating $2(90^\circ – \theta) = 180^\circ – 2\theta$. Since $\sin(180^\circ – 2\theta) = \sin(2\theta)$, calculated range values remain identical.