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 ā‰ˆ 9.81 m/s²).

Diagram of Projectile Motion (Interactive)

x (Horizontal) y (Vertical) O Īø R ymax vā‚€
Time of flight T: – s
Horizontal range R: – m
Maximum height ymax: – m

Here: Īø = angle of projection, vā‚€ = 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 vā‚€ sinĪø / g
  • Range: R = (v₀² sin 2Īø)/g
  • Maximum height: ymax = (v₀² sin²θ)/(2g)
  • Parametric path: x(t)=vā‚€ cosĪø Ā· t, y(t)=vā‚€ sinĪø Ā· t āˆ’ (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 Īø vā‚€ vā‚€ cosĪø vā‚€ sinĪø
v0x = vā‚€ cosĪø v0y = vā‚€ sinĪø ax = 0 ay = āˆ’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 = vā‚€ cosĪø, x = vā‚€ cosĪø Ā· t

Horizontal velocity

  • From diagram: cosĪø = ux/vā‚€ ⇒ ux = vā‚€ cosĪø
  • From first equation: v = u + a t, with ax=0 ⇒ vx = ux = vā‚€ cosĪø (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= āˆ’g and uy= vā‚€ sinĪø: y = vā‚€ sinĪø Ā· t āˆ’ (1/2) g t²
Vertical velocity (from 1st equation) v = u + a t vy = vā‚€ sinĪø āˆ’ g t
Speed–displacement relation (from 3rd equation) v² = u² + 2 a s With s ≔ y, ay= āˆ’g, uy= vā‚€ sinĪø: vy² = (vā‚€ sinĪø)² āˆ’ 2 g y
Quick consequences
  • Time to reach maximum height: tpeak = (vā‚€ sinĪø)/g (because vy=0 at the top).
  • Maximum height: ymax = (v₀² sin²θ)/(2g).
  • Range on level ground: R = (v₀² sin 2Īø)/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 cosĪø Ā· t
Horizontal displacement formula

Horizontal velocity (from the diagram)

cosĪø = ux / u Vx = u cosĪø

Also from first equation of motion

v = u + a t vx = ux + ax t
Since ax = 0 ⇒ vx = u cosĪø (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 = vā‚€ sinĪø Ā· t āˆ’ ½ g t²

Vertical velocity (1st formula)

vy = vā‚€ sinĪø āˆ’ g t

From third equation of motion

v² = u² + 2 a s vy² = (vā‚€ sinĪø)² + 2(āˆ’g) y vy² = (vā‚€ sinĪø)² āˆ’ 2 g y
At maximum height: vy = 0
⇒ ymax = (v₀² sin²θ)/(2g)

Reference Book: N/A

Author name: SIR H.A.Mwala Work email: biasharaboraofficials@gmail.com
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