10.3 Projectile Motion: Trajectory, Height, Flight Time & Range

Projectile Motion: Trajectory, Height, Flight Time & Range

Projectile motion is the two-dimensional motion of a body launched into space that continues moving under the sole influence of gravity and its own inertia. In the Intermediate Physics Part I syllabus and the UET ECAT entry test, projectile kinematics represents one of the most heavily tested subject areas. Candidates must master the parametric independence of orthogonal coordinate axes, trajectory geometry, range-height ratios, complementary angle relationships, and apex energy calculations.

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1. Principles of 2D Kinematics and Motion Independence

Galileo Galilei first proved that two-dimensional motion under constant acceleration can be analyzed as two simultaneous, mutually independent one-dimensional motions occurring along perpendicular axes.

Assumptions for Ideal Projectile Motion

1. Acceleration due to gravity $\mathbf{g}$ is constant in magnitude ($9.8\text{ m/s}^2 \approx 10\text{ m/s}^2$) and always points vertically downward.
2. Air resistance and aerodynamic drag are neglected.
3. The curvature and rotation of the Earth are ignored (valid for ranges much smaller than Earth's radius).

Initial Velocity Decomposition

Consider a projectile launched from origin $(0, 0)$ with an initial speed $v_0$ at an angle $\theta$ above the horizontal ground:

$\mathbf{v}_0 = v_{0x}\hat{\mathbf{i}} + v_{0y}\hat{\mathbf{j}} = (v_0\cos\theta)\hat{\mathbf{i}} + (v_0\sin\theta)\hat{\mathbf{j}}$

Kinematic Equations Along Coordinate Axes

Because horizontal acceleration is identically zero ($a_x = 0$), the horizontal component of velocity never changes throughout the entire flight.

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2. Derivation of the Trajectory Equation

To find the geometric path $y(x)$ followed by the projectile, eliminate time $t$ from the parametric equations of motion.

From the horizontal displacement equation:

$t = \frac{x}{v_0\cos\theta}$

Substituting this into the vertical displacement equation:

$y = (v_0\sin\theta)\left(\frac{x}{v_0\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$

Simplifying produces the Equation of Trajectory:

$y = x\tan\theta - \frac{g}{2v_0^2\cos^2\theta}x^2$

Using the trigonometric identity $\frac{1}{\cos^2\theta} = \sec^2\theta = 1 + \tan^2\theta$, the trajectory equation can be written as:

$y = x\tan\theta - \frac{g(1 + \tan^2\theta)}{2v_0^2}x^2$

Because this expression is of the quadratic form $y = Ax - Bx^2$ where $A = \tan\theta$ and $B = \frac{g}{2v_0^2\cos^2\theta} > 0$, the mathematical trajectory of an ideal projectile is a downward-opening parabola.

Alternative Form in Terms of Horizontal Range ($R$)

Factoring out $x\tan\theta$ and using $R = \frac{2v_0^2\sin\theta\cos\theta}{g}$:

$y = x\tan\theta\left(1 - \frac{g x}{2v_0^2\sin\theta\cos\theta}\right) = x\tan\theta\left(1 - \frac{x}{R}\right)$

This compact factored form allows rapid calculation of elevation $y$ at any position $x$ when range $R$ is known.

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3. Derivations of Core Projectile Parameters

1. Time to Reach Maximum Height ($t_{\text{up}}$)

At the apex of the parabolic path, the projectile stops ascending instantaneously, meaning vertical velocity vanishes ($v_y = 0$):

$0 = v_0\sin\theta - gt_{\text{up}} \implies t_{\text{up}} = \frac{v_0\sin\theta}{g}$

2. Total Time of Flight ($T$)

Because the trajectory over level ground is symmetric, the time of ascent equals the time of descent ($t_{\text{down}} = t_{\text{up}}$). Alternatively, setting vertical displacement $y = 0$:

$0 = (v_0\sin\theta)T - \frac{1}{2}gT^2 = T\left(v_0\sin\theta - \frac{1}{2}gT\right)$

Since $T \ne 0$ at landing:

$T = \frac{2v_0\sin\theta}{g} = 2t_{\text{up}}$

3. Maximum Height ($H$)

Using the third kinematic equation for vertical motion ($v_y^2 - v_{0y}^2 = 2a_y H$):

$0 - (v_0\sin\theta)^2 = 2(-g)H \implies H = \frac{v_0^2\sin^2\theta}{2g}$

Maximum height attains its absolute maximum when $\sin\theta = 1 \implies \theta = 90^\circ$ (vertical launch):

$H_{\text{max}} = \frac{v_0^2}{2g}$

4. Horizontal Range ($R$)

The horizontal range is the horizontal distance traveled during the total time of flight $T$:

$R = v_x \times T = (v_0\cos\theta)\left(\frac{2v_0\sin\theta}{g}\right) = \frac{v_0^2(2\sin\theta\cos\theta)}{g}$

Applying the double-angle identity $2\sin\theta\cos\theta = \sin 2\theta$:

$R = \frac{v_0^2\sin 2\theta}{g}$

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4. Key Relationships and Theorems in Projectile Motion

Maximum Horizontal Range

The horizontal range $R = \frac{v_0^2\sin 2\theta}{g}$ is maximized when $\sin 2\theta = 1$:

$2\theta = 90^\circ \implies \theta = 45^\circ$

$R_{\text{max}} = \frac{v_0^2}{g}$

Evaluating the maximum height attained at this optimum launch angle $\theta = 45^\circ$:

$H_{45^\circ} = \frac{v_0^2\sin^2(45^\circ)}{2g} = \frac{v_0^2(1/\sqrt{2})^2}{2g} = \frac{v_0^2}{4g} = \frac{R_{\text{max}}}{4}$

$R_{\text{max}} = 4H_{45^\circ}$

When a projectile is launched for maximum range, the maximum height it reaches is exactly one-fourth of its maximum range.

Complementary Angles Theorem

Two angles of projection $\theta_1$ and $\theta_2$ are complementary if $\theta_1 + \theta_2 = 90^\circ$. Let $\theta_1 = \theta$ and $\theta_2 = 90^\circ - \theta$:

$R(\theta_2) = \frac{v_0^2\sin[2(90^\circ - \theta)]}{g} = \frac{v_0^2\sin(180^\circ - 2\theta)}{g}$

Because $\sin(180^\circ - 2\theta) = \sin 2\theta$:

$R(\theta) = R(90^\circ - \theta)$

For a given initial launch speed, complementary launch angles yield identical horizontal ranges. For example, launch angles of $30^\circ$ and $60^\circ$, or $15^\circ$ and $75^\circ$, achieve identical landing ranges on level ground.

Relations for Complementary Projections

1. Ratio of Heights: $\frac{H_1}{H_2} = \frac{\sin^2\theta}{\sin^2(90^\circ - \theta)} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$
2. Sum of Heights: $H_1 + H_2 = \frac{v_0^2}{2g}(\sin^2\theta + \cos^2\theta) = \frac{v_0^2}{2g} = \frac{R_{\text{max}}}{2}$
3. Product of Flight Times: $T_1 T_2 = \left(\frac{2v_0\sin\theta}{g}\right)\left(\frac{2v_0\cos\theta}{g}\right) = \frac{2}{g}\left(\frac{v_0^2\sin 2\theta}{g}\right) = \frac{2R}{g}$

The Fundamental Height-to-Range Ratio

Dividing maximum height $H$ by horizontal range $R$:

$\frac{H}{R} = \frac{\frac{v_0^2\sin^2\theta}{2g}}{\frac{v_0^2(2\sin\theta\cos\theta)}{g}} = \frac{\sin^2\theta}{4\sin\theta\cos\theta} = \frac{1}{4}\tan\theta$

$\tan\theta = \frac{4H}{R}$

This universal relation is independent of launch speed $v_0$ and gravitational acceleration $g$.

• If $R = H$: $\tan\theta = 4 \implies \theta = \arctan(4) \approx 76^\circ$.
• If $R = 4H$: $\tan\theta = 1 \implies \theta = 45^\circ$.
• If $R = 4\sqrt{3}H$: $\tan\theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$.

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5. Instantaneous Velocity and Apex Energetics

Instantaneous Velocity Vector

At any elapsed time $t$ during flight:

$\mathbf{v}(t) = (v_0\cos\theta)\hat{\mathbf{i}} + (v_0\sin\theta - gt)\hat{\mathbf{j}}$

• Magnitude (Instantaneous Speed): $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(v_0\cos\theta)^2 + (v_0\sin\theta - gt)^2}$
• Direction Angle ($\phi$) with Horizontal: $\tan\phi = \frac{v_y}{v_x} = \frac{v_0\sin\theta - gt}{v_0\cos\theta}$

Kinetic Energy at the Highest Point

Let the launch kinetic energy be $K_0 = \frac{1}{2}m v_0^2$. At the highest point (apex), vertical velocity vanishes ($v_y = 0$), so the projectile's velocity is purely horizontal ($v_{\text{apex}} = v_0\cos\theta$):

$K_{\text{apex}} = \frac{1}{2}m v_x^2 = \frac{1}{2}m(v_0\cos\theta)^2 = \left(\frac{1}{2}mv_0^2\right)\cos^2\theta = K_0\cos^2\theta$

• At $\theta = 60^\circ$: $K_{\text{apex}} = K_0\cos^2(60^\circ) = K_0\left(\frac{1}{2}\right)^2 = \frac{1}{4}K_0$.
• At $\theta = 45^\circ$: $K_{\text{apex}} = K_0\cos^2(45^\circ) = K_0\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}K_0$.

Potential energy at the highest point is:

$U_{\text{apex}} = mgH = mg\left(\frac{v_0^2\sin^2\theta}{2g}\right) = \frac{1}{2}mv_0^2\sin^2\theta = K_0\sin^2\theta$

Total mechanical energy is strictly conserved:

$E_{\text{total}} = K_{\text{apex}} + U_{\text{apex}} = K_0(\cos^2\theta + \sin^2\theta) = K_0$

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6. Horizontal Projectile Launch from an Elevated Platform

Consider an object projected horizontally with speed $v_0$ from an elevation of height $h$ above level ground:

• Initial conditions: $v_{0x} = v_0, \quad v_{0y} = 0$
• Vertical fall: $h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$
• Horizontal range: $R = v_0 t = v_0\sqrt{\frac{2h}{g}}$
• Landing velocity components: $v_x = v_0, \quad v_y = -gt = -\sqrt{2gh}$
• Landing speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + 2gh}$

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Projectile Formulas Reference Table