Derive the equation of trajectory, the range, and the maximum height reached by a particle thrown at an obliqu

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Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.

The Precise Statement

g \approx 9.8 \, \text{m/s}^2

We break the motion into two independent components:

Horizontal motion: No acceleration ( $a_x = 0$ ). So horizontal velocity $v_x$ is constant.
Vertical motion: Constant downward acceleration ( $a_y = -g$ ). So vertical velocity $v_y$ changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

u_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

x = u_x t = (u \cos\theta) t

Vertical motion (constant acceleration $-g$ ):

v_y = u_y - gt = u \sin\theta - gt

y = u_y t - \frac{1}{2} g t^2 = (u \sin\theta) t - \frac{1}{2} g t^2

Key Results You Must Know

Important

Time of flight $T$ : total time the projectile stays in the air (until $y=0$ again).

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

Important

Maximum height $H$ : the highest vertical position reached (when $v_y = 0$ ).

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

Important

Range $R$ : the horizontal distance covered when it returns to launch height.

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

Note

The range is maximum when $\sin 2\theta = 1$ , i.e., $\theta = 45^\circ$ . For a given speed, $45^\circ$ gives the farthest throw.

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$ :

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

This is a parabola — the signature shape of projectile motion.

Common Mistake to Avoid …

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Step 1. Resolve the launch velocity. With launch speed $u$ at angle $\theta$ to the horizontal: $u_x=u\cos\theta$ (constant, since $a_x=0$ ), $u_y=u\sin\theta$ (decelerated by gravity, $a_y=-g$ ).

Step 2. Motion along each axis. Horizontal: $x=u_xt=(u\cos\theta)t$ . Vertical: $v_y=u_y-gt=u\sin\theta-gt$ ; $y=u_yt-\tfrac12gt^2=(u\sin\theta)t-\tfrac12gt^2$ .

Step 3. Trajectory equation. From the horizontal relation, $t=\dfrac{x}{u\cos\theta}$ . Substitute into the vertical-displacement relation:

y=u\sin\theta\cdot\frac{x}{u\cos\theta}-\frac12g\left(\frac{x}{u\cos\theta}\right)^2=x\tan\theta-\frac{g}{2u^2\cos^2\theta}x^2

This is a parabola in $x$ (inverted, opening downward).

Step 4. Maximum height. At the peak, $v_y=0$ . Using $v_y^2=u_y^2-2g\,s$ with $s=h_{max}$ : $0=(u\sin\theta)^2-2gh_{max} \Rightarrow h_{max}=\dfrac{u^2\sin^2\theta}{2g}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The value of acceleration at the highest point of the projectile path is$
(a) zero
(b) maximum
(c) minimum
(d) equal to g
›Reveal solutionSolution

Throughout projectile motion the acceleration equals g, so at the top it is also g. Answer (D).

In projectile motion (ignoring air resistance) the only force is gravity. Hence the acceleration is constant and equals g (about 9.8 m/s^2), pointing vertically downward, at all instants.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The maximum limit (range) of a gun is 3920 m. What is the velocity of shoot? (A) 200 m/s (B) 196 m/s (C) 100 m/s (D) 150 m/s$
›Reveal solutionSolution

The gun's muzzle speed is 196 m/s, found from the maximum-range formula of projectile motion.

The range of a projectile launched with speed $u$ at angle $\theta$ is $R = \dfrac{u^2\sin2\theta}{g}$ . The range is maximum when $\theta = 45^\circ$ (so $\sin2\theta = 1$ ), giving:

$R_{max} = \dfrac{u^2}{g}$

…

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CBSE 2024Set SET-AP55001 mark
$Q. The path of projectile motion is ________.$
›Reveal solutionSolution

The trajectory traced by a projectile (launched at an angle to the horizontal, under gravity alone) is a parabola.

For a projectile launched with initial speed u at angle θ, the horizontal motion has constant velocity u cos θ (no horizontal force, ignoring air resistance), so x = (u cos θ) t. The vertical motion is uniformly accelerated by gravity, so y = (u sin θ) t − ½gt^2.

Eliminating t (t = x / (u cos θ)) and substituting into the y-equation gives: …

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CBSE 2020Set ANNUAL1 mark
$Q. Answer in one word/one sentence: Which quantity is constant in projectile motion?$
›Reveal solutionSolution

Gravity gives the projectile a constant downward acceleration g, but no horizontal acceleration, so the horizontal velocity component vx stays fixed for the whole flight.

During projectile motion (ignoring air resistance), the only force acting is gravity, which acts vertically downward. This produces a constant downward acceleration g, changing the vertical velocity component continuously (vy = uy - gt). There is no horizontal force, so the horizontal acceleration is …

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Q.Derive the equation of trajectory, the range, and the maximum height reached by a particle thrown at an oblique angle θ\thetaθ with respect to the horizontal direction.

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Q.Derive the equation of trajectory, the range, and the maximum height reached by a particle thrown at an oblique angle $\theta$ with respect to the horizontal direction.

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