9. Mechanics — Dynamics

Projectile motion in vertical plane under gravity

Projectile motion in a vertical plane under a uniform gravitational field is governed by independent horizontal and vertical Newtonian kinematics, tracing a parabolic trajectory. This topic covers its time of flight, maximum height, horizontal range, and trajectory equation.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Independent Translations & Parabolic Trajectory Theorem

A particle projected from the origin $(0,0)$ with initial velocity $u$ at an angle of projection $\alpha$ to the horizontal under uniform acceleration due to gravity $g$ downwards satisfies the parametric equations of motion: $x = (u \cos\alpha)t$ and $y = (u \sin\alpha)t - \frac{1}{2}gt^2$. Eliminating the time parameter $t$ yields the Cartesian equation of trajectory: $$y = x \tan\alpha - \frac{g}{2u^2 \cos^2\alpha}x^2$$

📐 Geometric Interpretation

The trajectory represents a downward-opening parabola symmetric about its vertex, lying entirely in the vertical plane defined by the initial velocity vector and the vertical line of action of gravity.

⚙️ Algebraic & Structural Meaning

The Cartesian equation is a quadratic polynomial in $x$ of the form $y = ax - bx^2$, demonstrating that horizontal displacement and vertical displacement are coupled via a parabolic quadratic mapping.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Uniform Gravitational Field ($g = \text{constant}$)	`Entire trajectory duration $t \in [0, T]$`	Strictly Mandatory	If gravity varies with altitude or position, the equations of motion cease to be linear-quadratic and require non-uniform acceleration calculus.
Negligible Air Resistance (Zero Drag)	`Entire flight domain`	Mandatory for standard closed-form expressions	Presence of aerodynamic drag introduces a velocity-dependent deceleration term, making horizontal velocity non-constant and destroying the simple parabolic symmetry.
Flat Earth Approximation	`Spatial domain $x \in [0, R]$`	Essential for parallel gravity vectors	If the curvature of the Earth is considered, gravity vectors at different points are not parallel, modifying the effective range and trajectory profile.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Maximum Range & Projection Angle

$$R_{\max} = \frac{u^2}{g} \quad \text{at} \quad \alpha = 45^\circ$$

Rule: For a fixed initial speed $u$, the maximum horizontal range is always achieved at an angle of $45^\circ$ and equals twice the maximum height attained at that angle ($R_{\max} = 4H_{\max}$).

Application: Frequently asked direct conceptual and numerical questions in UP PGT and TGT Mathematics.

Complementary Angles of Projection

$$R_1 = R_2 \quad \text{when} \quad \alpha_1 + \alpha_2 = 90^\circ$$

Rule: Equal horizontal ranges are obtained for complementary angles of projection $\alpha$ and $90^\circ - \alpha$. Furthermore, the product of their times of flight is proportional to the range: $T_1 T_2 = \frac{2R}{g}$.

Application: Standard matching and calculation-based PYQ pattern in DSSSB PGT and KVS.

Angle of Elevation of Vertex & Range

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

Rule: The maximum height $H$ is related to the total horizontal range $R$ and angle of projection $\alpha$ via $H = \frac{1}{4} R \tan\alpha$.

Application: Directly used to solve inverse problems where $H$ and $R$ are given to find $\alpha$.

Velocity Vector Perpendicularity

$$\tan\alpha \cdot \tan\beta = -1 \implies \text{Orthonormal velocity change}$$

Rule: If a projectile is projected at angle $\alpha$ and later moves at angle $\beta$ to the horizontal, the time elapsed when the velocity vector is perpendicular to the initial velocity vector is given by $t = \frac{u}{g \sin\alpha}$.

Application: High-level conceptual PGT question testing vector dot products in dynamics.

Minimum Velocity for a Given Range

$$u_{\min} = \sqrt{gR}$$

Rule: The minimum initial velocity required to hit a target at a horizontal distance $R$ on level ground is $\sqrt{gR}$, achieved at an angle of $45^\circ$.

Radius of Curvature at the Highest Point

$$\rho_{\text{vertex}} = \frac{u^2 \cos^2\alpha}{g}$$

Rule: At the vertex (highest point), the normal acceleration is entirely due to gravity ($a_n = g$), so the radius of curvature simplifies to $\rho = \frac{v_x^2}{g} = \frac{u^2 \cos^2\alpha}{g}$.

Application: Use this result in the stated type of problem.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Range on Inclined Planes Confusion

Common Pitfall: Applying level-ground range formulas ($R = \frac{u^2 \sin 2\alpha}{g}$) directly to motion on an inclined plane.

Counterexample: Range up an incline of angle $\beta$: $$R_{\text{inc}} = \frac{u^2 [2 \sin(\alpha - \beta) \cos\alpha]}{\cos^2\beta}$$

Why it fails: The landing point is not at the same vertical level as the point of projection, altering the time of flight and kinematic components.

🛡️ Exam Defense: Always check whether the problem specifies motion on level ground or an inclined plane before selecting the formula.

⚠️ COMMON ERROR #02

Time of Flight vs Time to Maximum Height

Common Pitfall: Assuming total time of flight $T$ is equal to the time taken to reach maximum height $t_h$ in non-symmetrical setups.

Counterexample: For projection from a tower of height $h$ above ground: $T \neq 2t_h$.

Why it fails: Asymmetry in vertical displacement ($y(T) = -h$ instead of $0$) breaks the equal ascent/descent time symmetry.

🛡️ Exam Defense: Use the quadratic equation in time ($y(t) = ut - \frac{1}{2}gt^2$) to solve for $T$ rather than doubling the ascent time when starting/ending above/below projection level.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Level Ground Projectile	Zero air resistance, flat earth, horizontal launch surface	$T = \frac{2u\sin\alpha}{g}, H = \frac{u^2\sin^2\alpha}{2g}, R = \frac{u^2\sin 2\alpha}{g}$	Symmetric parabolic arc starting and ending at $y=0$	Keywords: 'level ground', 'horizontal plane'
Projectile on Inclined Plane	Uniform incline angle $\beta$, uniform gravity	$T_{\text{inc}} = \frac{2u \sin(\alpha-\beta)}{g \cos\beta}$	Parabola intersecting an inclined linear boundary	Keywords: 'plane inclined at angle $\beta$'
Projectile from Tower	Launch from height $h$ above ground	$h = ut \sin\alpha + \frac{1}{2}gt^2$	Asymmetric parabolic arc terminating below launch level	Keywords: 'tower of height h', 'cliff', 'dropped from top'
Envelope of Parabola (Safety Parabola)	Family of trajectories with constant initial speed $u$	$x^2 + y^2 + \frac{2u^2}{g}y - \frac{u^4}{g^2} = 0$	Boundary parabola enclosing all possible projectile paths	Keywords: 'envelope', 'maximum region reachable', 'safety parabola'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A projectile is fired with an initial velocity of $20\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$, find the maximum height attained and the total horizontal range.

(A) $H = 5\text{ m}, R = 20\sqrt{3}\text{ m}$(B) $H = 10\text{ m}, R = 10\sqrt{3}\text{ m}$(C) $H = 2.5\text{ m}, R = 40\sqrt{3}\text{ m}$(D) $H = 5\text{ m}, R = 10\sqrt{3}\text{ m}$

📚 Method A: Formal Textbook Derivation

Step 1: Identify parameters $u = 20$, $\alpha = 30^\circ$, $g = 10$. Step 2: Calculate $H = \frac{u^2 \sin^2\alpha}{2g} = \frac{400 \cdot (1/2)^2}{20} = \frac{100}{20} = 5\text{ m}$. Step 3: Calculate $R = \frac{u^2 \sin 2\alpha}{g} = \frac{400 \cdot \sin 60^\circ}{10} = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}\text{ m}$.

⚡ Method B: 25-Second Exam Shortcut

Use standard values: for $\alpha = 30^\circ$, $\sin\alpha = 1/2$, $\sin 2\alpha = \frac{\sqrt{3}}{2}$. Substitute directly into $H = \frac{u^2}{8g}$ and $R = \frac{\sqrt{3}u^2}{2g}$.

💡 Tip: Quick mental arithmetic with standard trigonometric ratios saves precious minutes in UP TGT exams.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the time of flight of a projectile is $10\text{ s}$ and its horizontal range is $500\text{ m}$, find the angle of projection $\alpha$. (Take $g = 10\text{ m/s}^2$)

(A) $30^\circ$(B) $45^\circ$(C) $60^\circ$(D) $75^\circ$

📚 Method A: Formal Textbook Derivation

Step 1: Write $T = \frac{2u \sin\alpha}{g} = 10 \implies u \sin\alpha = 50$. Step 2: Write $R = \frac{u^2 \sin 2\alpha}{g} = \frac{2u^2 \sin\alpha \cos\alpha}{g} = 500$. Step 3: Substitute $u \sin\alpha = 50$ into range formula: $\frac{2(50)u \cos\alpha}{10} = 500 \implies u \cos\alpha = 50$. Step 4: Thus $\tan\alpha = \frac{u \sin\alpha}{u \cos\alpha} = \frac{50}{50} = 1 \implies \alpha = 45^\circ$.

⚡ Method B: 25-Second Exam Shortcut

Note that $R = \frac{1}{2} g T^2 \tan\alpha$. Substituting values: $500 = \frac{1}{2} (10) (100) \tan\alpha = 500 \tan\alpha \implies \tan\alpha = 1 \implies \alpha = 45^\circ$.

💡 Tip: Remember the master relation $R = \frac{1}{2} g T^2 \tan\alpha$ for rapid problem-solving when $T$ and $R$ are given.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The equation of trajectory of a projectile is given by $y = \sqrt{3}x - \frac{g}{2}x^2$. The angle of projection $\alpha$ is:

A $30^\circ$

B $45^\circ$

C $60^\circ$

D $90^\circ$

Derivation & Analysis:

Comparing $y = x \tan\alpha - \frac{g}{2u^2 \cos^2\alpha}x^2$ with $y = \sqrt{3}x - \frac{g}{2}x^2$, we get $\tan\alpha = \sqrt{3}$, which implies $\alpha = 60^\circ$.

MCQ 2: If the maximum height of a projectile is equal to its horizontal range, the angle of projection $\alpha$ is given by:

A $\tan\alpha = 2$

B $\tan\alpha = 4$

C $\tan\alpha = 1$

D $\tan\alpha = \frac{1}{2}$

Derivation & Analysis:

Given $H = R$. We know $H = \frac{u^2 \sin^2\alpha}{2g}$ and $R = \frac{u^2 \sin 2\alpha}{g} = \frac{2u^2 \sin\alpha \cos\alpha}{g}$. Equating them gives $\frac{\sin^2\alpha}{2} = 2 \sin\alpha \cos\alpha \implies \tan\alpha = 4$.

MCQ 3: Two projectiles are projected with the same speed $u$ at angles $\alpha$ and $90^\circ - \alpha$. The ratio of their maximum heights $H_1 / H_2$ is:

A $\tan^2\alpha$

B $1$

C $\cot^2\alpha$

D $\tan\alpha$

Derivation & Analysis:

$H_1 = \frac{u^2 \sin^2\alpha}{2g}$ and $H_2 = \frac{u^2 \sin^2(90^\circ-\alpha)}{2g} = \frac{u^2 \cos^2\alpha}{2g}$. Thus $H_1 / H_2 = \tan^2\alpha$.

MCQ 4: At what point of the parabolic trajectory is the velocity minimum in magnitude?

A Point of projection

B Highest point (Vertex)

C Point of landing

D Midway of ascent

Derivation & Analysis:

The horizontal component of velocity $v_x = u \cos\alpha$ remains constant throughout, while the vertical component $v_y = 0$ at the highest point. Thus minimum speed occurs at the vertex.

MCQ 5: The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. The angle of projection is:

A $30^\circ$

B $45^\circ$

C $60^\circ$

D $75^\circ$

Derivation & Analysis:

Given $R = 4\sqrt{3} H$. Since $H = \frac{R \tan\alpha}{4}$, we have $R = 4\sqrt{3} \left(\frac{R \tan\alpha}{4}\right) \implies \tan\alpha = \frac{1}{\sqrt{3}} \implies \alpha = 30^\circ$.

MCQ 6: A particle is projected with velocity $u$ at angle $\alpha$. The magnitude of change in its momentum from projection to landing is:

A $2mu \sin\alpha$

B $2mu \cos\alpha$

C zero

D $mu \sin\alpha$

Derivation & Analysis:

Initial vertical momentum is $mu \sin\alpha$ upwards; final vertical momentum is $mu \sin\alpha$ downwards. The change in vertical momentum magnitude is $2mu \sin\alpha$ (horizontal momentum remains unchanged).

MCQ 7: The path of one projectile as observed from another projectile moving freely under gravity is always a:

A Parabola

B Circle

C Straight line

D Ellipse

Derivation & Analysis:

Since both projectiles experience the same acceleration due to gravity $g$ downwards, their relative acceleration is zero ($g - g = 0$). Hence, relative motion is uniform motion along a straight line.

MCQ 8: Assertion (A): Maximum horizontal range occurs at $\alpha = 45^\circ$. Reason (R): $\sin 2\alpha$ attains its maximum value of $1$ when $2\alpha = 90^\circ$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Range $R = \frac{u^2 \sin 2\alpha}{g}$ is maximized when $\sin 2\alpha = 1$, which gives $\alpha = 45^\circ$. Both assertion and reason are mathematically valid and linked.

MCQ 9: If a projectile is fired with velocity $u$ at angle $\alpha$ to an inclined plane of inclination $\beta$ (fired up the plane), the maximum range along the inclined plane is:

A $\frac{u^2}{g(1 + \sin\beta)}$

B $\frac{u^2}{g(1 - \sin\beta)}$

C $\frac{u^2}{g \cos^2\beta}$

D $\frac{u^2}{g(1 + \cos\beta)}$

Derivation & Analysis:

The maximum range up an inclined plane of angle $\beta$ is given by $R_{\max} = \frac{u^2}{g(1 + \sin\beta)}$, a standard derived formula in advanced dynamics.

Q10

MCQ 10: The radius of curvature of the trajectory at the point of projection for a projectile launched at angle $\alpha$ with speed $u$ is:

A $\frac{u^2 \cos^2\alpha}{g}$

B $\frac{u^2}{g \cos^3\alpha}$

C $\frac{u^2 \sin^2\alpha}{g}$

D $\frac{u^2}{g \sin^3\alpha}$

Derivation & Analysis:

Radius of curvature formula is $\rho = \frac{(1 + y'^2)^{3/2}}{|y''|}$. At $x=0$, $y' = \tan\alpha$ and $y'' = \frac{g}{u^2 \cos^2\alpha}$. Substituting gives $\rho = \frac{\sec^3\alpha}{(g / (u^2 \cos^2\alpha))} = \frac{u^2}{g \cos^3\alpha}$.

Common questions

In the absence of air resistance, there are no horizontal forces acting on the projectile. By Newton's First Law, horizontal acceleration is zero, maintaining a constant horizontal velocity component $v_x = u \cos\alpha$ throughout the flight.

Candidates frequently confuse level-ground formulas with inclined-plane formulas, or fail to account for launch heights that are non-zero (e.g., firing from a cliff or tower).

Memorize core relations such as $R_{\max} = \frac{u^2}{g}$, $H_{\max} = \frac{R_{\max}}{4}$ for $45^\circ$, and the complementary angle range property $R_1 = R_2$ when $\alpha_1 + \alpha_2 = 90^\circ$.

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