Projectile Motion Notes

Objectives: Projectile Motion Notes

Projectile Motion – Notes + Interactive Graph (SVG)

PROJECTILE MOTION :

Idea Projectile motion is the motion of an object that is thrown or projected into the air, moving under the influence of gravity alone (neglecting air resistance).

PROJECTILE :

Any object that is thrown or projected into the air and then moves under the action of gravity alone (e.g., a stone, ball, water drop, etc.).

Examples of projectiles

  • A stone thrown from the ground.
  • A bullet fired from a gun.
  • A football kicked into the air.
  • Water droplets from a fountain.
  • An athlete during the high jump.
Projectile motion combines two types of motion at the same time:
  • Horizontal motion – constant velocity (no acceleration horizontally).
  • Vertical motion – uniformly accelerated downward by gravity (g ≈ 9.81 m/s²).

Diagram of Projectile Motion (Interactive)

x (Horizontal) y (Vertical) O θ R ymax v₀
Time of flight T: – s
Horizontal range R: – m
Maximum height ymax: – m

Here: θ = angle of projection, v₀ = initial speed, R = horizontal range, ymax = maximum height.

Key point

The path (trajectory) of a projectile is a parabola.

Main assumptions in basic projectile problems

  • The resistance of air is neglected.
  • Acceleration due to gravity g is constant and acts downward.
  • The projectile does not lose speed due to air resistance.
  • Earth’s rotation and curvature are ignored (short-range motion).
Useful formulas (from the interactive model):
  • Time of flight: T = 2 v₀ sinθ / g
  • Range: R = (v₀² sin 2θ)/g
  • Maximum height: ymax = (v₀² sin²θ)/(2g)
  • Parametric path: x(t)=v₀ cosθ · t, y(t)=v₀ sinθ · t − (1/2) g t²
Projectile Motion – Horizontal vs Vertical (Derivations + Mini Demo)

PROJECTILE MOTION – Component Analysis

We split the motion into horizontal (no acceleration) and vertical (accelerated by gravity) parts. Take upward as positive and neglect air resistance.

Velocity components demo
x y θ v₀ v₀ cosθ v₀ sinθ
v0x = v₀ cosθ v0y = v₀ sinθ ax = 0 ay = −g

I) HORIZONTAL MOTION

Refers to: motion across the x-direction with no acceleration.

From second equation of motion s = ut + (1/2) a t² For horizontal motion, ax = 0, so: x = ux t and since ux = v₀ cosθ, x = v₀ cosθ · t

Horizontal velocity

  • From diagram: cosθ = ux/v₀ ⇒ ux = v₀ cosθ
  • From first equation: v = u + a t, with ax=0 ⇒ vx = ux = v₀ cosθ (constant).

II) VERTICAL MOTION

Refers to: upward and downward motion along y under gravity (take upward +ve).

Vertical displacement (from 2nd equation) y = uy t + (1/2) ay t² With ay= −g and uy= v₀ sinθ: y = v₀ sinθ · t − (1/2) g t²
Vertical velocity (from 1st equation) v = u + a t vy = v₀ sinθ − g t
Speed–displacement relation (from 3rd equation) v² = u² + 2 a s With s ≡ y, ay= −g, uy= v₀ sinθ: vy² = (v₀ sinθ)² − 2 g y
Quick consequences
  • Time to reach maximum height: tpeak = (v₀ sinθ)/g (because vy=0 at the top).
  • Maximum height: ymax = (v₀² sin²θ)/(2g).
  • Range on level ground: R = (v₀² sin 2θ)/g.
Projectile Motion – Notebook Style (as in the image)

From second equation of motion

s = ut + ½ a t² x = ux t + ½ ax t²
Here ax = 0 (no horizontal acceleration).
x = ux t
Now
x = u cosθ · t
Horizontal displacement formula

Horizontal velocity (from the diagram)

cosθ = ux / u Vx = u cosθ

Also from first equation of motion

v = u + a t vx = ux + ax t
Since ax = 0 ⇒ vx = u cosθ (constant).

II) VERTICAL MOTION

Refers to the movement of the projectile upwards and downwards under the influence of gravity.

Vertical displacement (2nd formula)

y = uy t + ½ ay t² y = v₀ sinθ · t − ½ g t²

Vertical velocity (1st formula)

vy = v₀ sinθ − g t

From third equation of motion

v² = u² + 2 a s vy² = (v₀ sinθ)² + 2(−g) y vy² = (v₀ sinθ)² − 2 g y
At maximum height: vy = 0
⇒ ymax = (v₀² sin²θ)/(2g)

Reference Book: N/A

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