Projectile Motion — From Basics to Tricky Problems

When you throw a ball at an angle, it follows a curved path — a parabola. Understanding why, and being able to calculate anything about that path, is what projectile motion is all about. Once you understand it truly, questions that seemed hard become straightforward. Let’s build this understanding from the ground up.

Key Terms & Definitions

Projectile: Any object that is given an initial velocity and then allowed to move under gravity alone (no thrust, no air resistance in the standard treatment).

Trajectory: The path followed by a projectile — a parabola.

Range (R): The horizontal distance covered by the projectile before hitting the ground.

Maximum Height (H): The greatest vertical height the projectile reaches above its launch point.

Time of flight (T): Total time the projectile spends in the air.

Angle of projection (θ): The angle at which the projectile is launched with respect to the horizontal.

Components of velocity: The key insight — a projectile has two independent motions simultaneously: horizontal (constant velocity) and vertical (uniformly accelerated by gravity).

The Core Idea — Independence of Motions

The deepest insight in projectile motion: horizontal and vertical motions are completely independent of each other. Gravity acts only vertically; it has absolutely no effect on horizontal motion.

Proof by thought experiment: Drop a ball vertically. Also throw an identical ball horizontally from the same height, at the same instant. Both balls hit the ground at exactly the same time. The thrown ball travels far horizontally, but its vertical motion is identical to the dropped ball. This experiment (demonstrable with a spring-loaded launcher and a falling target) proves independence.

Methods and Key Formulas

Breaking Velocity into Components

Initial velocity $u$ at angle $\theta$ :

Horizontal component: $u_x = u\cos\theta$ (constant throughout)
Vertical component: $u_y = u\sin\theta$ (decreasing on the way up, increasing on the way down)

At any time $t$ :

v_x = u\cos\theta \quad (\text{always constant})

v_y = u\sin\theta - gt

Equations of Motion

Horizontal (constant velocity):

x = u\cos\theta \cdot t

Vertical (uniform acceleration downward):

y = u\sin\theta \cdot t - \frac{1}{2}gt^2

Derived Formulas

Time of flight:

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

Maximum height:

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

Horizontal range:

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

Equation of trajectory:

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

(This is a parabola: $y = ax - bx^2$ form)

Range Maximisation

From $R = \frac{u^2\sin 2\theta}{g}$ : maximum range occurs when $\sin 2\theta = 1$ , i.e., $2\theta = 90°$ , i.e., $\theta = 45°$ .

R_{max} = \frac{u^2}{g} \quad \text{at } \theta = 45°

Complementary angles give equal range: If two projectiles are fired at angles $\theta$ and $(90° - \theta)$ with the same speed, they have the same range. This is because $\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90°-\theta))$ .

Example: 30° and 60° give equal range. 25° and 65° give equal range.

Velocity at Any Point

Magnitude: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(u\cos\theta)^2 + (u\sin\theta - gt)^2}$

At maximum height: $v_y = 0$ , so $v = u\cos\theta$ (only horizontal component remains).

Solved Examples

Easy — CBSE Level

Q: A ball is thrown horizontally with speed 20 m/s from a cliff 80 m high. Find (a) time to reach ground and (b) horizontal distance covered. (g = 10 m/s²)

This is “horizontal projectile” — launched horizontally, so $\theta = 0$ , $u_y = 0$ .

Vertical: $80 = \frac{1}{2} \times 10 \times t^2$

$t^2 = 16 \Rightarrow t = 4$ s

Horizontal: $x = 20 \times 4 = 80$ m

Answer: (a) 4 s, (b) 80 m

Medium — JEE Main Level

Q: A projectile has the same range when thrown at angles $\alpha$ and $\beta$ where $\alpha + \beta = 90°$ . If $H_1$ and $H_2$ are maximum heights for the two throws, prove that $H_1 + H_2 = R/4$ where R is the common range.

For $\theta = \alpha$ : $H_1 = \frac{u^2\sin^2\alpha}{2g}$

For $\theta = \beta = 90° - \alpha$ : $H_2 = \frac{u^2\sin^2(90°-\alpha)}{2g} = \frac{u^2\cos^2\alpha}{2g}$

H_1 + H_2 = \frac{u^2(\sin^2\alpha + \cos^2\alpha)}{2g} = \frac{u^2}{2g}

Range (same for both): $R = \frac{u^2\sin 2\alpha}{g} = \frac{2u^2\sin\alpha\cos\alpha}{g}$

\frac{R}{4} = \frac{u^2\sin\alpha\cos\alpha}{2g}

Hmm, $H_1 + H_2 = u^2/2g$ and $R/4 = u^2\sin\alpha\cos\alpha/2g$ . These are equal only when $\sin\alpha\cos\alpha = 1$ , which is impossible. Let me recheck: the correct identity is $R = 4\sqrt{H_1 H_2}$ .

Correct result: $R = 4\sqrt{H_1 H_2}$

\sqrt{H_1 H_2} = \sqrt{\frac{u^2\sin^2\alpha}{2g} \cdot \frac{u^2\cos^2\alpha}{2g}} = \frac{u^2\sin\alpha\cos\alpha}{2g} = \frac{u^2\sin 2\alpha}{4g} = \frac{R}{4}

Therefore $R = 4\sqrt{H_1 H_2}$ . ✓

Hard — JEE Advanced Level

Q: A particle is projected at angle 60° to horizontal with speed $u$ . Find the time when velocity makes 45° with horizontal.

Horizontal velocity: $v_x = u\cos 60° = u/2$

For velocity at 45° to horizontal: $v_y/v_x = \tan 45° = 1$ , so $v_y = v_x = u/2$ (considering downward phase — after maximum height)

After maximum height, $v_y = gt - u\sin 60° = gt - u\sqrt{3}/2$ (downward positive)

Setting $v_y = u/2$ : $gt - u\sqrt{3}/2 = u/2$

t = \frac{u(1 + \sqrt{3})}{2g}

If velocity makes 45° before maximum height (upward phase): $v_y/v_x = 1$ where $v_y = u\sin 60° - gt = u\sqrt{3}/2 - gt > 0$

$u\sqrt{3}/2 - gt = u/2 \Rightarrow t = \frac{u(\sqrt{3}-1)}{2g}$

Both answers are valid for different phases.

Exam-Specific Tips

JEE Main: Range formula $R = u^2\sin 2\theta/g$ and complementary angles giving equal range appear frequently. Integer-type questions often ask for time of flight or maximum height. Speed at a given height uses energy conservation: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mgh$ → $v^2 = u^2 - 2gh$ .

NEET: Simple projectile problems using the three basic formulae (T, H, R). Horizontal projectile (cliff problem) is common — remember $u_y = 0$ and $u_x = u$ . Maximum range at 45° is a conceptual MCQ.

CBSE Class 11: Derivations of T, H, R are required for 5-mark questions. Also the equation of trajectory (parabola equation). Start from $x = u\cos\theta \cdot t$ → $t = x/(u\cos\theta)$ , substitute into the $y$ equation.

Common Mistakes to Avoid

Mistake 1: Using $g$ in horizontal equations. Gravity has NO horizontal component. Horizontal velocity is constant throughout. Any formula involving $g$ applies only to vertical motion.

Mistake 2: Forgetting that at maximum height, vertical velocity is zero — but horizontal velocity is NOT zero. The ball is still moving! Its speed at maximum height is $u\cos\theta$ , not zero.

Mistake 3: For horizontal projectile (thrown horizontally), using $u\sin\theta$ when $\theta = 0$ gives $u_y = 0$ . This is correct — initial vertical velocity is zero for horizontal throws. But the vertical velocity builds up due to gravity. Many students forget to include gravity in the vertical direction for horizontal projectiles.

Mistake 4: Confusing time of flight for half the parabola vs full. Time to reach maximum height = $T/2 = u\sin\theta/g$ . Total time of flight = $T = 2u\sin\theta/g$ .

Mistake 5: Applying range formula $R = u^2\sin 2\theta/g$ when the launch and landing points are at DIFFERENT heights. This formula is only valid when the projectile returns to the same height it was launched from. For different heights, use the full kinematic equations.

Practice Questions

Q1. A ball is projected at 30° with speed 40 m/s. Find the range and maximum height. (g = 10 m/s²)

Range: $R = \frac{(40)^2 \sin 60°}{10} = \frac{1600 \times \sqrt{3}/2}{10} = 80\sqrt{3} \approx 138.6$ m

Maximum height: $H = \frac{(40)^2 \sin^2 30°}{2 \times 10} = \frac{1600 \times 1/4}{20} = 20$ m

Q2. At what angle should a ball be projected to get maximum range? What is the maximum range if initial speed is 20 m/s?

Maximum range at $\theta = 45°$ . $R_{max} = \frac{u^2}{g} = \frac{400}{10} = \textbf{40 m}$

Q3. A stone is thrown horizontally from a 125 m cliff with velocity 10 m/s. Find time of flight and landing distance from base of cliff. (g = 10 m/s²)

Vertical: $125 = \frac{1}{2}(10)t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5$ s. Horizontal: $x = 10 \times 5 = 50$ m from base.

Q4. Two balls are projected at angles 25° and 65° with the same speed. Compare their ranges.

Since $25° + 65° = 90°$ (complementary angles), their ranges are equal. $\sin(2 \times 25°) = \sin 50° = \sin(180° - 50°) = \sin 130° = \sin(2 \times 65°)$ .

Q5. Find the time at which velocity of a projectile is perpendicular to its initial velocity. (Initial speed u, angle θ)

At time t, velocity components: $v_x = u\cos\theta$ , $v_y = u\sin\theta - gt$ .

For perpendicular: $\vec{u} \cdot \vec{v} = 0 \Rightarrow u\cos\theta \cdot u\cos\theta + u\sin\theta(u\sin\theta - gt) = 0$

$u^2\cos^2\theta + u^2\sin^2\theta - ugt\sin\theta = 0$

$u^2 = ugt\sin\theta \Rightarrow t = \frac{u}{g\sin\theta}$

Q6. A projectile’s range equals twice its maximum height. Find the angle of projection.

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

$2\sin\theta\cos\theta = \sin^2\theta \Rightarrow 2\cos\theta = \sin\theta \Rightarrow \tan\theta = 2$

$\theta = \arctan(2) \approx 63.4°$

Q7. Velocity of projectile at half the maximum height. (Initial speed $u$ , angle $\theta = 60°$ )

$H = \frac{u^2\sin^2 60°}{2g} = \frac{3u^2}{8g}$

At $h = H/2 = \frac{3u^2}{16g}$ , using energy: $v^2 = u^2 - 2gh = u^2 - 2g \times \frac{3u^2}{16g} = u^2 - \frac{3u^2}{8} = \frac{5u^2}{8}$

$v = u\sqrt{5/8} = \frac{u\sqrt{10}}{4}$

FAQs

Q: Why is the trajectory a parabola and not a circle or ellipse?

Because horizontal velocity is constant and vertical acceleration is constant ( $g$ ). When $x = u_x t$ and $y = u_y t - \frac{1}{2}gt^2$ , eliminating $t$ gives $y = \frac{u_y}{u_x}x - \frac{g}{2u_x^2}x^2$ — a quadratic in $x$ , which is a parabola. A circle would require a force always perpendicular to velocity (like a string), which gravity is not.

Q: Does air resistance affect projectile motion?

Yes, significantly for real projectiles. Air resistance reduces range, reduces maximum height, makes the trajectory asymmetric (steeper descent than ascent), and makes the angle for maximum range less than 45°. In JEE and NEET, we ignore air resistance unless specifically stated.

Q: Is the horizontal velocity really constant?

In the ideal model (no air resistance), yes — there’s no horizontal force, so horizontal velocity never changes. Air resistance opposes motion, so it has a horizontal component that reduces $v_x$ over time. For school and JEE problems, assume ideal conditions.

Q: What’s the velocity at the highest point?

Only the horizontal component remains: $v = u\cos\theta$ . The direction is horizontal. The vertical component is zero at the highest point.

Q: Can I use energy conservation for projectile problems?

Yes! Energy is conserved (no friction). $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgh$ gives velocity at any height $h$ . This is often faster than kinematic equations for height-velocity problems.