Projectile Motion Notes

Objectives: Projectile Motion Notes

Projectile Motion - Horizontal Projection

Projectile Motion – Key Quantities

  • Maximum Height: Hmax = u² sin²θ / (2g)
  • Range: R = (u² sin2θ)/g
  • Velocity of projectile at any point: Vx = u cosθ (constant)
    Vy = u sinθ − g t

Horizontal Projection (θ = 0°) from Height (h)

Consider diagram:

h x O Vx Vy

Equations of Motion

  • Horizontal motion: Vx = u (constant) x = u t
  • Vertical motion: Vy = − g t y = − ½ g t² (negative indicates downward displacement)

Time of Flight

From equation: y = −½ g t² If vertical displacement = h, then:

h = ½ g t² T = √(2h/g)
Projectile from Height

Projectile Motion – From a Height

Using equation:

Vy² = Uy² + 2 a y

For vertical displacement from height h:

0 = 0 − 2 g h t = √(2h / g)

👉 Time taken by projectile to hit the ground:

T = √(2h / g)

Horizontal Distance

x = Ux × t x = U cosθ × √(2h/g)

Projectile Fired at an Angle θ Above Horizontal (from height h)

Consider the diagram:

y x O θ Vx Vy

Initial Components

  • Ux = U cosθ
  • Uy = U sinθ
  • ay = −g

Time to Reach Top

Tup = (U sinθ) / g

Height Above Launch

H = (U² sin²θ) / (2g)
Projectile Motion – Range & Velocity at Impact

Height Above Ground

Hmax = h + (U² sin²θ) / (2g)

Time of Flight

Consider vertical motion:

H = h + U sinθ · t − ½ g t²

At ground (y=0):

0 = h + U sinθ · t − ½ g t²

Rearrange:

½ g t² − U sinθ · t − h = 0

Solve quadratic:

t = (U sinθ ± √((U sinθ)² + 2gh)) / g

Since time cannot be negative:

T = (U sinθ + √((U sinθ)² + 2gh)) / g

Horizontal Range (R)

R = Ux · T = U cosθ · T R = U cosθ (U sinθ + √((U sinθ)² + 2gh)) / g

Velocity at Impact

At final impact:

  • Horizontal component: Vx = U cosθ (constant)
  • Vertical component: Vy = U sinθ − gT
Vimpact = √(Vx² + Vy²) α = tan⁻¹(Vy / Vx)
Projectile Motion Notes

Projectile Fired Below Horizontal From Top

U h R

Let initial velocity be U at angle θ below horizontal.

  • Horizontal component: Ux = U cosθ
  • Vertical component (downward): Uy = U sinθ

Equation of Motion

y = Uy·t + ½ g t²

Time of Flight

h = U sinθ·t + ½ g t² T = (U sinθ + √((U sinθ)² + 2gh)) / g

Range

R = Ux · T = U cosθ · T

Projectile on an Inclined Plane

A projectile is fired on a slope instead of flat ground. The landing point lies on the inclined surface.

U α θ

Definitions

  • α = angle of inclination of the plane above horizontal
  • (θ − α) = effective angle with the inclined plane
  • U = initial velocity of projectile
Projectile on Inclined Plane - Extended Notes

Projectile Motion on an Inclined Plane (Detailed)

Resolving Initial Velocity

  • Along the plane: Ux = U cos(θ − α)
  • Perpendicular to plane: Uy = U sin(θ − α)

Components of Acceleration

  • Along x-axis (along plane): ax = − g sinα
  • Along y-axis (perpendicular to plane): ay = − g cosα

Diagram

U α θ−α P

Equations of Motion (Inclined Coordinates)

x = U cos(θ − α)·t − ½ g sinα·t² y = U sin(θ − α)·t − ½ g cosα·t²

Note: At landing point, y = 0

Formulas

1. Time of Flight From y = U sin(θ − α)·t − ½ g cosα·t² At landing point y = 0 ⇒ T = (2 U sin(θ − α)) / (g cosα)

This is the time of flight along inclined plane.

2. Range Along the Plane From x = U cos(θ − α)·t − ½ g sinα·t² Substitute t = T: R = [U² sin(2(θ − α))] / [g cosα]
Projectile on Inclined Plane - Page 3

Range on Inclined Plane

From previous expression for range along the plane:

R = (U² / g cosα) · sin(2θ − α) / (1 + sinα)

For maximum range:

Condition: sin(2θ − α) = 1

⇒ 2θ − α = 90° ⇒ θ = 45° + α/2

Expression for Maximum Range

Rmax = U² / [ g (1 + sinα) ]

Important Notes

  • θ − α = firing angle with the inclined plane.
  • When θ = 45° + α/2, the range along the incline is maximum.

Projectile Motion of Two Bodies Thrown Simultaneously

Techniques to solve:

  1. Identify the common parameters (e.g., distance, time, height, etc.).
  2. Formulate equations for these parameters for both projectiles.
Projectile Motion - Page 4

Projectile Motion of Two Bodies

Techniques:

  1. Equate the two equations and solve unknowns.
  2. If the bodies collide, their horizontal velocities and time of flight are compared (when projected from the same horizontal plane).
  3. If the bodies collide at the same time, their horizontal velocities and vertical motions must also match (when projected from the same vertical plane).

Worked Example

A ball is thrown with a speed of U = 17 m/s at an angle of 58° above the horizontal. Assume it returns at the same horizontal level. Find:

  1. Time of flight
  2. The range
  3. Maximum height
  4. Time taken to reach maximum height

Solutions

1. Time of Flight

T = (2U sinθ) / g T = (2 × 17 × sin 58°) / 9.8 T ≈ 2.9 s

2. Range

R = (U² sin 2θ) / g R = (17² × sin(116°)) / 9.8 R ≈ 26.5 m

3. Maximum Height

Hmax = (U² sin²θ) / (2g) Hmax = (17² × sin²58°) / (2 × 9.8) Hmax ≈ 11.5 m

4. Time to Reach Maximum Height

tup = (U sinθ) / g tup = (17 × sin 58°) / 9.8 tup ≈ 1.45 s

Final Results:

  • Time of Flight ≈ 2.9 s
  • Range ≈ 26.5 m
  • Maximum Height ≈ 11.5 m
  • Time to reach maximum height ≈ 1.45 s

Reference Book: N/A

Author name: SIR H.A.Mwala Work email: biasharaboraofficials@gmail.com
#MWALA_LEARN Powered by MwalaJS #https://mwalajs.biasharabora.com
#https://educenter.biasharabora.com

:: 1::

⬅ ➡