Projectile Motion

Independence of Motion

Projectile motion is a two-dimensional motion where horizontal and vertical components act independently under the sole influence of gravity.

$$x = v_x \times t$$

The horizontal distance depends only on constant horizontal velocity and time.

$x$=Horizontal displacement(m)

$v_x$=Horizontal velocity component (constant)(m/s)

$t$=Time elapsed(s)

No air resistance

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$v_x$ remains constant throughout the flight — the projectile has zero horizontal acceleration.

Horizontal Component: No horizontal force acts on the projectile (ignoring air resistance), so $a_x = 0$ and horizontal velocity remains constant.

Vertical Component: The object accelerates downwards at $g = 9.8$ m/s² regardless of its horizontal speed.

Stroboscopic Proof: A dropped ball and a horizontally projected ball hit the ground at the same time if released from the same height — their vertical motions are identical.

Dimensional Check: $[x] = [v_x][t] = (\text{m/s})(\text{s}) = \text{m}$ confirms displacement units.

The vertical displacement of any projectile follows the same equation as free fall, with the initial vertical velocity determining whether it rises first or falls directly.

$$y = v_{iy}t - \frac{1}{2}gt^2$$

Vertical position as a function of time, combining initial upward velocity and gravitational pull.

$y$=Vertical displacement (positive upward)(m)

$v_{iy}$=Initial vertical velocity ($v_i \sin\theta$)(m/s)

$g$=Acceleration due to gravity(m/s²)

Horizontal launch ($\theta = 0$)

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$v_{iy} = 0$, so $y = -\frac{1}{2}gt^2$ — pure free fall.

At time of flight ($y = 0$ again)

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The projectile returns to launch height, giving $t = \frac{2v_{iy}}{g}$.

Horizontal Launch: When thrown horizontally from a cliff, $v_{iy} = 0$ and the fall time depends only on height: $t = \sqrt{\frac{2h}{g}}$.

Sign Convention: Taking upward as positive, gravity is $-g$, so the vertical velocity decreases on the way up and increases (in magnitude) on the way down.

Oblique Projectile Components

When launched at an angle $\theta$, the initial velocity $v_i$ must be resolved into its orthogonal components using trigonometry.

$$v_{fy} = v_i \sin \theta - gt$$

The vertical velocity changes linearly over time due to constant gravitational acceleration.

$v_{fy}$=Vertical velocity at time $t$(m/s)

$v_i \sin \theta$=Initial vertical component of velocity(m/s)

$g$=Acceleration due to gravity(m/s²)

At peak height ($v_{fy} = 0$)

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Time to peak: $t_{peak} = \frac{v_i \sin\theta}{g}$, which is exactly half the total flight time.

At landing (same level)

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$v_{fy} = -v_i \sin\theta$ — equal magnitude but opposite direction to launch.

Resolution of Vectors: $v_{ix} = v_i \cos \theta$ (constant throughout) and $v_{iy} = v_i \sin \theta$ (changes due to gravity).

Constant Velocity: $v_x$ remains $v_i \cos \theta$ at every point on the parabolic trajectory.

Symmetry: The time taken to reach the peak is exactly half the total time of flight for ground-to-ground launches.

Speed at Landing: For same-level launches, the landing speed equals the launch speed: $v_f = v_i$. Only the direction changes.

Eliminating time from the horizontal and vertical displacement equations yields the trajectory equation — a parabola relating $y$ directly to $x$.

$$y = x \tan\theta - \frac{g x^2}{2 v_i^2 \cos^2\theta}$$

The path equation showing vertical position as a function of horizontal position — a downward-opening parabola.

$x \tan\theta$=The straight-line path the projectile would follow without gravity

$\frac{g x^2}{2 v_i^2 \cos^2\theta}$=The gravitational 'drop' below the straight-line path

$\theta = 45^\circ$

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$\tan\theta = 1$ and $\cos^2\theta = 0.5$, giving $y = x - \frac{gx^2}{v_i^2}$.

Derivation: Substitute $t = \frac{x}{v_i \cos\theta}$ from the horizontal equation into the vertical equation $y = v_i \sin\theta \cdot t - \frac{1}{2}gt^2$.

Parabolic Shape: The $x^2$ term confirms the trajectory is a parabola opening downward.

Dimensional Verification: $[x \tan\theta] = \text{m}$ and $\left[\frac{g x^2}{v_i^2 \cos^2\theta}\right] = \frac{(\text{m/s}^2)(\text{m}^2)}{(\text{m/s})^2} = \text{m}$ — both terms have correct dimensions.

Key Parameters: Height, Time, and Range

The maximum height is reached when the vertical velocity component becomes zero at the peak of the trajectory.

$$H = \frac{v_i^2 \sin^2 \theta}{2g}$$

Maximum vertical displacement, derived from $v_{fy}^2 = v_{iy}^2 - 2gH$ with $v_{fy} = 0$.

$H$=Maximum height above launch point(m)

$v_i \sin\theta$=Initial vertical velocity component(m/s)

$g$=Gravitational acceleration(m/s²)

$\theta = 90^\circ$ (vertical throw)

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$H = \frac{v_i^2}{2g}$ — maximum possible height for a given speed.

$\theta = 45^\circ$

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$H = \frac{v_i^2}{4g}$ and the range is $R = 4H$.

Proportionality Shortcut: $H \propto v_i^2$ — if speed triples, height increases by factor 9.

Limiting Case: At $\theta = 0^\circ$, $H = 0$ — a horizontally launched projectile gains no height above its launch point.

The time of flight is the total duration from launch to landing at the same elevation, and equals twice the time to reach peak height.

$$T = \frac{2 v_i \sin \theta}{g}$$

Total flight time for a ground-to-ground projectile, derived from the symmetry of the parabolic path.

$T$=Total time of flight(s)

$v_i \sin\theta$=Initial vertical velocity component(m/s)

$g$=Gravitational acceleration(m/s²)

$\theta = 90^\circ$

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$T = \frac{2v_i}{g}$ — maximum flight time (straight up and down).

$\theta = 30^\circ$ vs $\theta = 60^\circ$

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The $60^\circ$ projectile stays airborne longer ($\sin 60 > \sin 30$) even though both have the same range.

Proportionality Shortcut: $T \propto v_i$ (linear) — unlike $H$ and $R$ which scale as $v_i^2$.

Half-time to Peak: Time to reach maximum height is $t_{peak} = T/2 = \frac{v_i \sin\theta}{g}$.

Dimensional Check: $[T] = \frac{(\text{m/s})}{(\text{m/s}^2)} = \text{s}$ — dimensions are consistent.

The horizontal range of a projectile depends on both the initial speed and the launch angle through the double-angle sine function.

$$R = \frac{v_i^2 \sin 2\theta}{g}$$

The horizontal distance covered during the entire flight, found by multiplying $v_x$ by $T$.

$R$=Horizontal range(m)

$v_i$=Initial launch speed(m/s)

$\sin 2\theta$=Sine of double the launch angle

$\theta = 45^\circ$

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$\sin 90^\circ = 1$ gives maximum range: $R_{max} = \frac{v_i^2}{g}$.

Complementary angles ($\theta$ and $90^\circ - \theta$)

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$\sin 2\theta = \sin(180^\circ - 2\theta)$, so both angles give the same range.

Maximum Range: At $\theta = 45^\circ$, the range is maximized because $\sin 90^\circ = 1$.

Complementary Angle Rule: Angles like $30^\circ$ and $60^\circ$ produce identical ranges — the steeper angle gives more height but less horizontal speed, and vice versa.

Height-Range Ratio: At $\theta = 45^\circ$, $R = 4H$. This shortcut avoids recalculating from scratch.

Ground-to-Ground Only: This formula assumes launch and landing occur at the same vertical level.

Instantaneous Velocity

At any point during flight, the projectile's speed is found by combining its constant horizontal and changing vertical velocity components using the Pythagorean theorem.

$$V = \sqrt{v_{fx}^2 + v_{fy}^2}$$

The magnitude of the resultant velocity vector at any instant.

$V$=Instantaneous speed (magnitude of velocity)(m/s)

$v_{fx}$=Horizontal velocity ($= v_i \cos\theta$, constant)(m/s)

$v_{fy}$=Vertical velocity at time $t$ ($= v_i \sin\theta - gt$)(m/s)

At the peak

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$v_{fy} = 0$, so $V_{min} = v_{fx} = v_i \cos\theta$ — the minimum speed during flight.

At launch and landing (same level)

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$V = v_i$ — the speed is the same at both points (only direction changes).

Minimum Speed: Occurs at the peak where only horizontal velocity remains: $V_{min} = v_i \cos\theta$.

Direction of Velocity: The angle $\phi$ with the horizontal at any time is $\tan\phi = \frac{v_{fy}}{v_{fx}}$.

Velocity vs Speed: At the peak, velocity is purely horizontal (not zero!) — only the vertical component is zero.

Real-world Ballistics

Real-world projectiles, such as a ballistic missile, deviate from ideal parabolic paths due to aerodynamic forces.

$$R_{max} = \frac{v_i^2}{g}$$

The theoretical maximum range in a vacuum at $45^\circ$ — an upper bound for real-world performance.

$R_{max}$=Maximum possible range (vacuum, flat Earth)(m)

$v_i$=Initial launch speed(m/s)

Air friction present

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The actual range is significantly less than $R_{max}$ and the optimal angle is less than $45^\circ$.

Ideal vs Actual: Air drag causes the path to be shorter and asymmetric — the descent is steeper than the ascent.

Aerodynamic Forces: At high speeds, air friction can exceed gravitational force, making ideal equations completely unrealistic.

Ballistic Flight: An unpowered, unguided flight moving solely due to inertia and gravity.