Projectile Motion Notes

Objectives: Projectile Motion Notes

Projectile Motion Notes

Vertical Velocity

From: vy = uy + ay t vy = u sinθ − g t

Resultant velocity of projectile

V = √(vx² + vy²) V = √((u cosθ)² + (u sinθ − g t)²)

After time t:

vx = u cosθ vy = u sinθ − g t ⇒ V = √((u cosθ)² + (u sinθ − g t)²)

Direction angle of motion

θ = tan⁻¹ (vy / vx)
x y u
Note:
  • When projectile is thrown upward, g is taken as negative.
  • When projectile descends (free fall), g is positive.

Parameters of Projectile Motion

  • Angle of projection
  • Trajectory (path)
  • Time of flight
  • Maximum height
  • Time to reach max height
  • Range
  • Velocity of projectile

Angle of Projection

The angle that the projectile makes with the horizontal when it is launched.

Projectile Motion - Horizontal Direction

Horizontal Direction

It determines the trajectory (path), range, maximum height, and the time of flight of the projectile.

Horizontal Displacement

x = ux t + ½ ax t² ax = 0 x = u cosθ · t t = x / (u cosθ)

Vertical Displacement

y = uy t + ½ g t² y = u sinθ · t − ½ g t²

Trajectory

The curved path followed by a projectile.

x y trajectory

Trajectory Equation

When we consider both horizontal and vertical motion:

y = x tanθ − (g x²) / (2 u² cos²θ)

Notes

  • Small θ → longer horizontal range but smaller height.
  • Large θ → higher path but shorter horizontal distance.
Projectile Motion - Time of Flight

Equation of Projectile (general)

y = (tanθ)·x − (g / (2 u² cos²θ)) · x²

Time of Flight (T)

The total time taken by the projectile to move up, reach the highest point, and return back to the ground.

T = (2 u sinθ) / g
Note: Time of flight represents total time in air.

Derivation (from vertical displacement)

y = u sinθ · t − ½ g t²

At time of flight, body returns to the ground, so y = 0:

0 = u sinθ · T − ½ g T² T = (2 u sinθ) / g

Time to Reach Maximum Height (Tup)

Tup = (u sinθ) / g

At maximum height, vertical velocity = 0.

T = 2 Tup
Projectile Motion - Maximum Height and Range

Maximum Height (Hmax)

It is the highest vertical point a projectile reaches during its flight.

From Newton’s third equation:

v² = u² − 2 g h

At maximum height, vertical velocity v = 0, and u = u sinθ.

0 = (u sinθ)² − 2 g Hmax Hmax = (u² sin²θ) / (2 g)
Condition for Maximum Height:
At maximum height, θ = 90° → sinθ = 1.

Horizontal Range (R)

It is the total horizontal distance covered by a projectile to its landing point.

From horizontal displacement:

x = u cosθ · t

At landing point, t = T (time of flight).

R = u cosθ · T

But T = (2 u sinθ)/g

R = (u cosθ · 2 u sinθ) / g R = (u² sin2θ) / g
Note: Since 2 sinθ cosθ = sin2θ, the range formula simplifies neatly.
Projectile Motion - Special Cases

Condition for Maximum Range

From range formula: R = (u² sin2θ)/g

For maximum value of R → sin2θ = 1.

2θ = 90° → θ = 45° Rmax = (u²)/g

Gravity Notes

  • Time to reach highest point: Tup = u/g
  • Total Time of flight: T = 2u/g
  • Maximum height: Hmax = u² / (2g)

Special Cases of Projectile

1. Vertical Projection (θ = 90°)

Projectile is thrown straight upward.

  • Horizontal range = 0 (no horizontal displacement)
  • Vertical motion: slows down until it stops, then accelerates back downward under gravity.

2. Projectile Fired at an Angle with Vertical

90°−θ H R

Now the angle of projection is measured from vertical. We can plug this into equations of trajectory, time of flight, etc.

Reference Book: N/A

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