Projectile motion

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2015, 2018, 2021, 2022, 2023)
• Priority tier: T2
• Marks (count): 10 (2), 13 (1), 15 (2)
• Average solve time: ~9 min
• Difficulty mix: medium 4, easy 1
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom appears in Section B in 5 of the last 13 years, making it one of the more reliable Dynamics questions in Paper 1. Every UPSC question reduces to writing the parabolic trajectory in range-factored form $y = (\tan\alpha/R)\,x(R-x)$ and then reading off the constraint; once you recognise which archetype you are in, the algebra follows mechanically. The four archetypes (geometric constraints, locus of vertices, perpendicular-direction proof, range from height) have been set at least once each since 2015 — breadth is thin, depth is shallow, scoring potential is high.

Minimum Theory

Standard setup. A particle is projected from the origin $O$ with speed $u$ at angle $\alpha$ above the horizontal. Taking $Ox$ horizontal and $Oy$ vertically upward, the equations of motion under gravity $g$ are

$x = u\cos\alpha\cdot t,\qquad y = u\sin\alpha\cdot t - \tfrac{1}{2}g\,t^2.$

Eliminating $t$ gives the parabolic trajectory

$y = x\tan\alpha - \frac{g\,x^2}{2u^2\cos^2\!\alpha}.$

Standard results.

Range-factored form (key trick). Using $R = u^2\sin 2\alpha/g$ and $\tan\alpha = 2H/R\cdot\ldots$, the trajectory rewrites as

$\boxed{y = \frac{\tan\alpha}{R}\,x(R-x).}$

This form is zero at $x=0$ and $x=R$ automatically; it never contains $u$ or $g$ separately, only their combination $R$. When the problem gives $R$ as data (or the range is geometrically fixed), substituting the constraint point $(d,h)$ immediately gives $\tan\alpha$.

Two angles for the same range. For given $u$ the range $R = u^2\sin 2\alpha/g$ is the same for complementary angles $\alpha_1$ and $\alpha_2 = \pi/2 - \alpha_1$.

Trajectory with OY vertically downward. If OY points downward (gravity acts in the positive $y$-direction), the trajectory equation changes sign in the gravity term:

$y = x\tan\alpha + \frac{g\,x^2}{2u^2}\sec^2\!\alpha.$

This sign flip is the only change; all other kinematics are standard.

Figure: Standard projectile parabola with apex at $(R/2,\,H)$, range $R$, time of flight $T$, and the range-factored form $y = (\tan\alpha/R)\,x(R-x)$.

Question Archetypes

projectile-constraints (2 question(s); 2015, 2018)

Find launch speed $u$ and/or elevation $\alpha$ so a projectile satisfies one or two geometric constraints

Recognition Cues

• The question specifies a wall, cone, or target that the projectile must just clear or land on.
• Two unknowns ($u$ and $\alpha$) appear — you need two equations.
• Either the range is fixed geometrically (so use range-factored form) or the apex is pinned (use $H$ formula).
• Phrases: “just clears”, “grazes the apex”, “strikes at a point”, “horizontal range $R$”.

Solution Template

1. Set up coordinates so the launch point is a convenient origin; identify the landing point and any intermediate constraint point.
2. Write the range-factored trajectory $y = (\tan\alpha/R)\,x(R-x)$ if $R$ is known or can be expressed geometrically; otherwise write the full trajectory $y = x\tan\alpha - gx^2\sec^2\!\alpha/(2u^2)$.
3. Impose each constraint (e.g., the point the projectile must pass through) to get equations in $u$ and $\alpha$.
4. Solve the system — typically divide one equation by the other to get $\tan\alpha$, then back-substitute for $u^2$.
5. State the answer clearly: exact form preferred ($u = \sqrt{\ldots}$, $\alpha = \arctan(\ldots)$).

Worked Example 1

2015 Paper 1, 2015-P1-Q7b (13 marks)

A particle is projected from the foot of a cone whose semi-vertical angle is $30°$ and height $h$. The particle just grazes the apex of the cone and lands at the diametrically opposite point on the base circle. Find the initial speed $u$ and angle of projection $\alpha$.

Setup. The cone has height $h$ and semi-vertical angle $30°$, so the base radius is $r = h\tan 30° = h/\sqrt{3}$. Place the origin at the launch point (the foot of the cone on the near edge of the base). The apex is at $(r, h) = (h/\sqrt{3},\,h)$ and the landing point (diametrically opposite base edge) is at $(2r, 0) = (2h/\sqrt{3},\,0)$.

Key observation. The apex lies at horizontal distance $h/\sqrt{3}$ from the launch point, which is exactly half the horizontal range $R = 2h/\sqrt{3}$. A point at $x = R/2$ on a parabola is the apex of the parabola (the trajectory’s highest point). Therefore the projectile’s vertex coincides with the cone apex — this is both the “grazing” condition and the apex-of-trajectory condition.

Two equations. The apex of the trajectory has height $H = u^2\sin^2\!\alpha/(2g)$ and lies at $y=h$: $u^2\sin^2\!\alpha = 2gh \qquad\cdots (1)$

The range is $R = 2h/\sqrt{3}$ and $R = u^2\sin 2\alpha/g$, so $u^2 \cdot 2\sin\alpha\cos\alpha = 2gh/\sqrt{3}$: $u^2\sin\alpha\cos\alpha = \frac{gh}{\sqrt{3}} \qquad\cdots (2)$

Divide (1) by (2): $\tan\alpha = \frac{2gh}{gh/\sqrt{3}} = 2\sqrt{3}.$

$\boxed{\alpha = \arctan(2\sqrt{3})}$

Back-substitute into (1). From $\tan\alpha = 2\sqrt{3}$: $\sin^2\!\alpha = 12/13$. Then $u^2 = 2gh\cdot(13/12)$, giving

$\boxed{u = \sqrt{\frac{13gh}{6}}.}$

Verification. $\tan\alpha = 2\sqrt{3} \approx 3.46 > \sqrt{3}$, which confirms the launch direction is steeper than the cone slant ($\tan 60° = \sqrt{3}$), so the particle genuinely clears the exterior of the cone.

Worked Example 2

2018 Paper 1, 2018-P1-Q5e (10 marks)

A projectile is fired with speed $u$ at elevation $\alpha$ and just clears a wall of height $h$ at horizontal distance $d$ from the point of projection. If the horizontal range on level ground is $R$, find $\tan\alpha$.

Apply the range-factored form. The trajectory is $y = (\tan\alpha/R)\,x(R-x)$. At $x = d$ the height equals $h$:

$h = \frac{\tan\alpha}{R}\cdot d(R-d).$

Solve for $\tan\alpha$:

$\boxed{\tan\alpha = \frac{hR}{d(R-d)}.}$

No further computation is needed. The formula is valid as long as $d < R$ (the wall is within the range).

Common Traps

• Apex at $x=R/2$, not at the constraint point. In the cone problem the constraint point happens to be the apex because $x_{\text{apex}} = R/2$ — this is a coincidence you must verify, not assume. Confirm by checking the horizontal position of the grazing point equals $R/2$.
• Forgetting the factor $(R-d)$ in the denominator. Writing $\tan\alpha = h/d$ instead of $hR/[d(R-d)]$ is the most common slip in the wall problem.
• Direction of the launch vs. cone slant. After finding $\alpha$, verify $\tan\alpha > \tan(\text{cone slant angle})$ to confirm the flight is outside (not through) the cone.

projectile-locus (1 question(s); 2021)

Find the locus traced by the vertices of a family of projectile parabolas as the projection angle varies with fixed launch speed

Recognition Cues

• Fixed $u$; angle $\alpha$ (or $\theta$) is a parameter that varies over $(0°, 90°)$.
• “Find the locus of the vertex” or “find the curve traced by the highest point.”
• Parametric elimination: write $(X,Y)$ for the vertex in terms of $\alpha$, eliminate $\alpha$.

Solution Template

1. Write the vertex coordinates as functions of $\alpha$: $X = \frac{u^2\sin 2\alpha}{2g} = \frac{u^2\sin\alpha\cos\alpha}{g}, \qquad Y = \frac{u^2\sin^2\!\alpha}{2g}.$
2. Set $\phi = 2\alpha$ and express $X = (u^2/2g)\sin\phi$, $Y = (u^2/2g)(1-\cos\phi)/2$ — recognise as a trigonometric parametrisation of an ellipse or circle.
3. Eliminate $\phi$ using $\sin^2\!\phi + \cos^2\!\phi = 1$ to get the Cartesian locus.
4. State the centre and semi-axes; check $Y\ge 0$ (vertex is always above the $x$-axis).

Worked Example

2021 Paper 1, 2021-P1-Q8b (15 marks)

A particle is projected with velocity $u = 4\sqrt{g}$ at various angles from a fixed point. Find the locus of the vertices of the paths.

Vertex coordinates. With $u^2 = 16g$:

$X = \frac{u^2\sin 2\theta}{2g} = \frac{16g\sin 2\theta}{2g} = 8\sin 2\theta,$ $Y = \frac{u^2\sin^2\!\theta}{2g} = \frac{16g\sin^2\!\theta}{2g} = 8\sin^2\!\theta = 4(1-\cos 2\theta).$

Eliminate $\theta$. Let $\phi = 2\theta \in (0,\pi)$. Then $X = 8\sin\phi$ and $Y-4 = -4\cos\phi$, so

$\frac{X^2}{64} + \frac{(Y-4)^2}{16} = \sin^2\!\phi + \cos^2\!\phi = 1.$

$\boxed{\frac{x^2}{64}+\frac{(y-4)^2}{16}=1.}$

This is an ellipse centred at $(0,4)$ with semi-major axis $8$ (horizontal) and semi-minor axis $4$ (vertical). Only the upper half ($y\ge 0$, corresponding to $\phi\in(0,\pi)$) is traced, but UPSC accepts the full ellipse equation.

Common Traps

• Off-by-two error in $Y$. It is easy to write $Y = u^2\sin 2\theta/(2g)$ (the range formula) for the vertex height. The correct formula is $Y = u^2\sin^2\!\theta/(2g)$.
• Missing the centre shift. The ellipse is centred at $(0,4)$, not the origin. Writing $x^2/64 + y^2/16 = 1$ is wrong.
• Only upper half. Strictly $\theta\in(0°,90°)$ traces the upper half of the ellipse; if the question asks for “locus” state the full ellipse.

projectile-property-proof (1 question(s); 2022)

Prove a geometric property of trajectories, typically involving the Vieta relations for two complementary projection directions through a given point

Recognition Cues

• Fixed $u$ (often expressed as $u^2 = 2gh$ for algebraic neatness); a specific point $(x,y)$ is reachable by two different angles.
• Asked to prove the two directions are perpendicular, or that one direction bisects a specified angle.
• Axes may be unconventional: OY vertically downward (gravity in positive $y$-direction).

Solution Template

1. Write the trajectory equation for the given axis convention; substitute $u^2 = 2gh$.
2. Rearrange as a quadratic in $\tan\alpha$ (the two roots are $\tan\alpha_1$ and $\tan\alpha_2$ for the two angles reaching point $(x,y)$).
3. Apply Vieta’s formulae: product of roots $= \tan\alpha_1\tan\alpha_2$; sum of roots $= \tan\alpha_1 + \tan\alpha_2$.
4. For the perpendicularity condition: two lines at angles $\alpha_1, \alpha_2$ to the horizontal are perpendicular iff $\tan\alpha_1\tan\alpha_2 = -1$. Set the Vieta product equal to $-1$ and simplify.
5. For the bisector condition: identify the bisector direction using the half-angle identity $\tan(\beta/2) = (1-\cos\beta)/\sin\beta$ where $\beta = \angle POX$.

Worked Example

2022 Paper 1, 2022-P1-Q5d (10 marks)

A particle is projected from $O$ with speed $u$ where $u^2 = 2gh$. Axes: $OX$ horizontal, $OY$ vertically downward. From $O$, two directions of projection reach the same point $P = (x,y)$ on the trajectory. Show that (i) if the two directions are perpendicular, the locus of $P$ is $x^2 = 2hy$; (ii) if one direction bisects the angle $POX$, find that direction.

Trajectory with OY downward. With gravity in the $+y$ direction: $y = x\tan\alpha + \frac{gx^2}{2u^2}\sec^2\!\alpha.$ Substitute $u^2 = 2gh$ and $\sec^2\!\alpha = 1 + \tan^2\!\alpha$: $y = x\tan\alpha + \frac{gx^2}{4gh}(1+\tan^2\!\alpha) = x\tan\alpha + \frac{x^2}{4h}(1+\tan^2\!\alpha).$

Rearrange as a quadratic in $t = \tan\alpha$:

$\frac{x^2}{4h}\,t^2 + x\,t + \frac{x^2}{4h} - y = 0. \qquad\cdots(\star)$

Vieta’s formulae for the two roots $\tan\alpha_1, \tan\alpha_2$: $\tan\alpha_1 + \tan\alpha_2 = -\frac{4h}{x}, \qquad \tan\alpha_1\tan\alpha_2 = 1 - \frac{4hy}{x^2}.$

Part (i) — Perpendicular directions. The two projection directions are perpendicular iff $\tan\alpha_1\tan\alpha_2 = -1$: $1 - \frac{4hy}{x^2} = -1 \implies \frac{4hy}{x^2} = 2 \implies \boxed{x^2 = 2hy.}$

Part (ii) — Bisector of $\angle POX$. Let $\beta = \angle POX$, so $\tan\beta = y/x$. From $x^2 = 2hy$ on the locus: $\tan\beta = x/(2h)$.

The bisector direction has $\tan\alpha = \tan(\beta/2)$. Using the half-angle identity: $\tan\!\tfrac{\beta}{2} = \frac{1-\cos\beta}{\sin\beta}.$

Express $\sin\beta$ and $\cos\beta$ from $\tan\beta = x/(2h)$: $\sin\beta = x/\sqrt{x^2+4h^2}$, $\cos\beta = 2h/\sqrt{x^2+4h^2}$.

Substitute and note $x^2+4h^2 = 2hy + 4h^2$ (using $x^2=2hy$): $\tan\!\tfrac{\beta}{2} = \frac{\sqrt{x^2+4h^2}-2h}{x}.$

Verify this is a root of $(\star)$: since $x^2=2hy$, both roots of $(\star)$ are real and their Vieta product is $-1$, confirming they are perpendicular; the bisector root is the one with smaller absolute slope.

Common Traps

• Sign of the gravity term. With OY downward the trajectory is $y = x\tan\alpha \mathbf{+} gx^2\sec^2\!\alpha/(2u^2)$ — both terms are positive. Using the standard upward formula (minus sign) gives the wrong quadratic and the wrong Vieta product.
• Perpendicularity condition. Two directions at angles $\alpha_1, \alpha_2$ to the horizontal are perpendicular iff $\tan\alpha_1\tan\alpha_2 = -1$, not $= 1$. The product equals $-1$ because slopes of perpendicular lines satisfy $m_1 m_2 = -1$.
• Half-angle denominator. The identity $\tan(\beta/2)=(1-\cos\beta)/\sin\beta$ is valid for $\beta\in(0,\pi)$, which covers all acute $\angle POX$.

projectile-range (1 question(s); 2023)

Compare the range $R_2$ of a projectile launched from height $h$ above the landing level with the sea-level range $R_1$

Recognition Cues

• Two scenarios: same speed $v$, same angle $\theta$, but one launched from height $h$ (Range $R_2$) and one from the landing level (Range $R_1$).
• Asked to “show $R_2 > R_1$” and find the ratio $(R_2 - R_1):R_1$.
• Solving a quadratic in $t$ (time of flight from the elevated position) is unavoidable; take the positive root.

Solution Template

1. Write $R_1 = v^2\sin 2\theta/g$ (standard range formula).
2. For $R_2$: solve the vertical displacement equation $-h = v\sin\theta\cdot t - \tfrac{1}{2}g\,t^2$ (landing is $h$ below launch). Take the positive root: $t = \frac{v\sin\theta + \sqrt{v^2\sin^2\!\theta + 2gh}}{g}.$
3. Write $R_2 = v\cos\theta\cdot t$.
4. Form $R_2 - R_1$ and show it is positive.
5. Express the ratio $(R_2 - R_1)/R_1$ in simplified form.

Worked Example

2023 Paper 1, 2023-P1-Q6b (15 marks)

A particle is projected with speed $v$ at elevation $\theta$ from the top of a cliff of height $h$. The sea-level range is $R_1$ (as if projected from sea level). Let $R_2$ be the actual range. Prove $R_2 > R_1$ and find the ratio $(R_2 - R_1):R_1$.

Sea-level range. $R_1 = \frac{v^2\sin 2\theta}{g} = \frac{2v^2\sin\theta\cos\theta}{g}.$

Time of flight from height $h$. The particle falls a net height $h$ below the launch point, so $-h = v\sin\theta\cdot t - \tfrac{1}{2}g\,t^2$, i.e., $\tfrac{1}{2}g\,t^2 - v\sin\theta\cdot t - h = 0$. Positive root:

$t = \frac{v\sin\theta + \sqrt{v^2\sin^2\!\theta + 2gh}}{g}.$

Range from height. $R_2 = v\cos\theta\cdot t = \frac{v\cos\theta}{g}\!\left(v\sin\theta + \sqrt{v^2\sin^2\!\theta+2gh}\right).$

Prove $R_2 > R_1$.

$R_2 - R_1 = \frac{v\cos\theta}{g}\!\left(v\sin\theta + \sqrt{v^2\sin^2\!\theta+2gh}\right) - \frac{2v^2\sin\theta\cos\theta}{g}$ $= \frac{v\cos\theta}{g}\!\left(\sqrt{v^2\sin^2\!\theta+2gh} - v\sin\theta\right).$

Since $h>0$, $\sqrt{v^2\sin^2\!\theta+2gh} > \sqrt{v^2\sin^2\!\theta} = v\sin\theta$. Thus $R_2 - R_1 > 0$, i.e., $R_2 > R_1$. $\square$

Find the ratio.

$\frac{R_2-R_1}{R_1} = \frac{\dfrac{v\cos\theta}{g}\!\left(\sqrt{v^2\sin^2\!\theta+2gh}-v\sin\theta\right)}{\dfrac{2v^2\sin\theta\cos\theta}{g}} = \frac{\sqrt{v^2\sin^2\!\theta+2gh}-v\sin\theta}{2v\sin\theta}.$

Factor $v\sin\theta$ out of the square root:

$\boxed{(R_2-R_1):R_1 = \frac{1}{2}\!\left(\sqrt{1+\frac{2gh}{v^2\sin^2\!\theta}}-1\right):1.}$

Common Traps

• Sign when landing below launch. If launch is at height $h$ above the landing level and $y$ is measured upward from launch, then the landing condition is $y = -h$, giving $\tfrac{1}{2}gt^2 - v\sin\theta\cdot t - h = 0$. Using $+h$ on the right gives a shorter (wrong) flight time.
• Choosing the wrong root. The quadratic in $t$ has two roots; the physical one (landing after launch) is always the positive root — equivalently, the root with the $+$ sign before the square root.
• Simplification of the ratio. The clean form requires factoring $v\sin\theta$ from inside the square root: $\sqrt{v^2\sin^2\!\theta+2gh} = v\sin\theta\cdot\sqrt{1+2gh/(v^2\sin^2\!\theta)}$. This step is easy to skip, leaving a messy answer that may not match the expected boxed form.

Marks-Aware Writing

For a 10-mark question (e.g., 2018-P1-Q5e, 2022-P1-Q5d): One clean derivation suffices. Write the trajectory in the appropriate form, substitute the constraint, and box the answer. Show one intermediate line of algebra explicitly — examiners cannot award method marks for hidden steps.

For a 13-mark question (e.g., 2015-P1-Q7b): Set up coordinates clearly (1–2 lines), state both equations labelled (1) and (2), divide to get $\tan\alpha$, back-substitute for $u^2$, take the square root, and state both answers boxed. Add the brief verification that the launch angle exceeds the cone’s slant angle.

For a 15-mark question (e.g., 2021-P1-Q8b, 2023-P1-Q6b): The method must be broken into legible stages: vertex parametrisation and elimination (locus problem), or time-of-flight derivation and range comparison (height problem). State the final answer in boxed display math. For the ratio problem, explicitly write the inequality step before calling the proof complete.

OY-downward problems (2022). State the axis convention at the top of your solution. Write the trajectory equation with the $+$ sign (downward gravity) to signal to the examiner you know the convention. Any answer written with the standard $-$ sign will be marked wrong even if the final locus is accidentally correct.

Practice Set

• 2024-P1-Q6b (15 m) — — projectile over a wall; find maximum range for given wall dimensions; use range-factored form and optimise over $\alpha$
• 2020-P1-Q5c (10 m) — — complementary angles and the range formula; verify sum/difference identities hold
• 2016-P1-Q5e (10 m) — — projectile constraints: given two conditions derive the launch parameters