Projectile Motion

Master horizontal and oblique projection for JEE Physics

Prerequisites

Before studying this topic, make sure you understand:

What is Projectile Motion?

A projectile is any object that is thrown into the air and moves under the influence of gravity alone (no air resistance).

Think Spider-Man!
Remember when Spider-Man swings between buildings in No Way Home or Across the Spider-Verse? When he releases his web at the highest point of his swing and flies through the air — that’s projectile motion! His path traces a perfect parabola until he shoots another web.

Key Characteristics:

  • Horizontal motion: uniform velocity (no acceleration)
  • Vertical motion: uniformly accelerated ($a = g$ downward)
  • Path: Parabola
Independence Principle

The horizontal and vertical motions are completely independent. What happens in x doesn’t affect y, and vice versa.

Movie Connection: In Pathaan (2023), when Shah Rukh Khan jumps from helicopters or between buildings, his horizontal speed stays constant while gravity pulls him down — classic projectile motion!

Interactive Demo

Explore projectile motion with this full-featured simulator. Adjust parameters and observe how they affect the trajectory:

How to Use This Animation

Controls:

  • Initial Velocity (u): Drag the slider to set launch speed (10-100 m/s)
  • Launch Angle: Set the angle of projection (0-90 degrees)
  • Gravity Toggle: Switch between Earth (g = 9.8 m/s squared) and Moon (g = 1.6 m/s squared)
  • Play/Pause: Start or pause the animation
  • Reset: Return projectile to starting position

What to Observe:

Blue arrow (vx): Horizontal velocity stays CONSTANT throughout the flight

Green arrow (vy): Vertical velocity decreases going up, becomes zero at apex, then increases downward

Red dashed arrow (v): Resultant velocity (vector sum of vx and vy)

Green dot at apex: Maximum height where vy = 0

Orange range marker: Horizontal distance covered

Try These Experiments:

  1. Set angle to 45 degrees and note the range. Then try 30 degrees and 60 degrees - they have the SAME range!
  2. Enable “Show complementary angle” to see both trajectories simultaneously
  3. Switch to Moon gravity - notice how much farther the projectile travels
  4. At any angle, observe that vx never changes while vy constantly decreases

Types of Projectile Motion

1. Horizontal Projection

Object is thrown horizontally from a height $h$ with velocity $u$.

Pushpa 2 Cliff Scene
In Pushpa 2: The Rule, when objects (or people!) are thrown horizontally off a cliff, they don’t fall straight down — they travel forward while falling. The horizontal throw determines how FAR they land, while gravity determines how FAST they fall!

Initial Conditions:

  • $u_x = u$ (horizontal)
  • $u_y = 0$ (no vertical component)

Equations:

DirectionMotion TypeEquations
Horizontal (x)Uniform$x = ut$
Vertical (y)Accelerated$y = \frac{1}{2}gt^2$, $v_y = gt$

Key Formulas:

$$\boxed{T = \sqrt{\frac{2h}{g}}}$$ $$\boxed{R = u\sqrt{\frac{2h}{g}} = uT}$$ $$\boxed{v_{final} = \sqrt{u^2 + 2gh}}$$
Time of Flight
Time of flight depends only on height, not on horizontal velocity!

2. Oblique Projection (Ground to Ground)

Object is thrown at angle $\theta$ with horizontal, with initial velocity $u$.

xyOuθu cos θu sin θHIGHEST POINTv = u cos θ (horizontal only)HR (Range)v = u cos θv (final)t = 0t = T/2t = TgPARABOLICTRAJECTORYINITIALVELOCITY uKey Formulas:H = u²sin²θ / 2gR = u²sin2θ / gT = 2u sinθ / gMax Range at θ = 45°R_max = u²/gKey Insight:Horizontal: Uniform motionVertical: Free fall (a = g)Both motions independent!

Initial Components:

  • $u_x = u\cos\theta$
  • $u_y = u\sin\theta$

At any time t:

  • $x = (u\cos\theta)t$
  • $y = (u\sin\theta)t - \frac{1}{2}gt^2$
  • $v_x = u\cos\theta$ (constant)
  • $v_y = u\sin\theta - gt$

Key Formulas for Oblique Projection

Time of Flight

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

Time to reach maximum height = $\frac{T}{2} = \frac{u\sin\theta}{g}$

Maximum Height

$$\boxed{H_{max} = \frac{u^2\sin^2\theta}{2g}}$$

Horizontal Range

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

Equation of Trajectory

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

This is a parabola (quadratic in x).

Alternative form: $y = x\tan\theta\left(1 - \frac{x}{R}\right)$

Important Properties

Maximum Range

Range is maximum when $\sin 2\theta = 1$, i.e., $\theta = 45°$

$$R_{max} = \frac{u^2}{g}$$
Jawan Train Scene
In Jawan (2023), remember when SRK throws objects or jumps between train cars? For maximum horizontal distance, 45° is the magic angle! That’s why athletes in long jump or javelin throw aim for approximately this angle.

Complementary Angles

For angles $\theta$ and $(90° - \theta)$:

  • Same range (since $\sin 2\theta = \sin 2(90° - \theta)$)
  • Different heights and times
AngleRangeHeightTime
$30°$$R$$H_1$$T_1$
$60°$$R$ (same)$H_2 > H_1$$T_2 > T_1$

Velocity at Any Point

At height $h$:

$$v = \sqrt{u^2 - 2gh}$$

At any time:

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

Velocity Direction

$$\tan\alpha = \frac{v_y}{v_x} = \frac{u\sin\theta - gt}{u\cos\theta}$$

where $\alpha$ is angle with horizontal.

Common Mistake

Don’t confuse:

  • Height at time $t$: $y = u_y t - \frac{1}{2}gt^2$
  • Maximum height: $H_{max} = \frac{u^2\sin^2\theta}{2g}$

Special Cases

Projection from a Height

If projectile is launched from height $h$ above ground:

Time of flight: Solve $h + (u\sin\theta)T - \frac{1}{2}gT^2 = 0$

$$T = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}$$

Projection on Inclined Plane

For a plane inclined at angle $\alpha$:

  • Range along plane: $R = \frac{2u^2\sin(\theta - \alpha)\cos\theta}{g\cos^2\alpha}$
  • Maximum range: at $\theta = \frac{\pi}{4} + \frac{\alpha}{2}$

Summary Table

QuantityHorizontal ProjectionOblique Projection
Initial $v_x$$u$$u\cos\theta$
Initial $v_y$$0$$u\sin\theta$
Time of flight$\sqrt{\frac{2h}{g}}$$\frac{2u\sin\theta}{g}$
Max height$\frac{u^2\sin^2\theta}{2g}$
Range$u\sqrt{\frac{2h}{g}}$$\frac{u^2\sin 2\theta}{g}$

Practice Problems

Problem 1

A ball is thrown horizontally from a 45 m high cliff with speed 20 m/s. Find time of flight and range. (g = 10 m/s²)

Solution: $T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{10}} = \sqrt{9} = $ 3 s

$R = uT = 20 \times 3 = $ 60 m

Problem 2

A projectile is launched at 30° with 40 m/s. Find range, max height, and time of flight. (g = 10 m/s²)

Solution: $T = \frac{2u\sin\theta}{g} = \frac{2 \times 40 \times 0.5}{10} = $ 4 s

$H_{max} = \frac{u^2\sin^2\theta}{2g} = \frac{1600 \times 0.25}{20} = $ 20 m

$R = \frac{u^2\sin 2\theta}{g} = \frac{1600 \times \sin 60°}{10} = \frac{1600 \times 0.866}{10} ≈ $ 138.6 m

Problem 3

For what angle of projection is range equal to maximum height?

Solution: $R = H_{max}$ $\frac{u^2\sin 2\theta}{g} = \frac{u^2\sin^2\theta}{2g}$ $2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$ $4\cos\theta = \sin\theta$ $\tan\theta = 4$ $\theta = \tan^{-1}(4) ≈ $ 76°


Within Kinematics

Connected Chapters

Math Connections