Lesson 2: Projectile Motion | PHY 2.4

Lesson at a Glance | He Tirohanga Whakamua

Do NowWhich hits the ground first: a dropped bullet vs a fired bullet?10 min

Vector ResolutionResolving initial velocity into

v_{x} = v \cos θ

v_{y} = v \sin θ

15 min

Trajectory MathCalculating peak height, flight time, and horizontal range15 min

Portfolio EntryWrite Section 2: Projectile Trajectory & Vector Resolution Map10 min

Exit VerificationExplain why vertical velocity is zero at the peak5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

That perpendicular vectors are independent: horizontal velocity ( $v_{x}$ ) remains constant ( $a_{x} = 0 {ms}^{- 2}$ ), while vertical motion experiences constant acceleration due to gravity ( $a_{y} = - 9.8 {ms}^{- 2}$ ).
How to resolve initial velocity at launch angle $θ$ :
• $v_{x} = v_{i} \cos θ$
• $v_{y i} = v_{i} \sin θ$
How to calculate Peak Height ( $v_{y} = 0 {ms}^{- 1}$ ), Total Flight Time ( $t_{t o t a l} = 2 t_{p e a k}$ ), and Horizontal Range ( $d_{x} = v_{x} t_{t o t a l}$ ).

Students will demonstrate

By solving a complete angled projectile trajectory scenario (e.g. rugby ball kicked at $20 {ms}^{- 1}$ at 35°).
By completing Section 2 of their Level 2 Physics Mechanics Mastery Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Independence of Vectors Prompt:

"If you drop a tennis ball from a 1.8m cliff height, it takes 0.6 seconds to hit the ground. If you throw a second tennis ball horizontally off the same cliff at 20 ms⁻¹, how long does it take to hit the ground?"

Unpack: Exactly 0.6 seconds! Horizontal motion has zero effect on vertical acceleration due to gravity ( $g = 9.8 {ms}^{- 2}$ ). Both balls fall vertically at the exact same rate, demonstrating the independence of perpendicular motion vectors.

Anatomy of an Angled Projectile Trajectory (15 min)

1. Horizontal Component ( $x$ -axis)

$v_{x} = v_{i} \cos θ$ (Constant throughout flight)
$a_{x} = 0 {ms}^{- 2}$ (No horizontal forces in vacuum)
Range: $d_{x} = v_{x} \times t_{t o t a l}$

2. Vertical Component ( $y$ -axis)

$v_{y i} = v_{i} \sin θ$ (Initial upward velocity)
$a_{y} = - 9.8 {ms}^{- 2}$ (Constant gravity acceleration)
Peak: $v_{y, p e a k} = 0 {ms}^{- 1}$

Resolving the Launch Velocity (15 min)

1. Split the launch into two independent motions

A projectile launched at speed $v$ and angle $θ$ has $v_{x} = v \cos θ$ across and $v_{y} = v \sin θ$ up. The two are independent: what happens horizontally does not affect what happens vertically.

2. Only one component changes

Ignoring air resistance, $v_{x}$ stays constant for the whole flight because no horizontal force acts. $v_{y}$ changes at $g = 9.8$ m s⁻² downward, reaching zero at the peak. That is why the object is still moving at the top of its arc, and why a dropped and a horizontally-fired object hit the ground together.

Do this now: resolve a 20 m s⁻¹ launch at 35° into its two components, then state which one you would use to find time of flight.

📁 Physics Mechanics Portfolio — Section 2: Projectile Trajectory & Vectors

Students open their Level 2 Physics Portfolio and complete Section 2:

Section 2 Requirements:

1. Annotated Parabolic Trajectory Diagram: Draw a launch vector at 40° showing $v_{x}$ constant arrows and $v_{y}$ shrinking to 0 at the peak and expanding downwards.

2. Multi-Step Trajectory Calculation: Solve for $v_{x}$ , $v_{y i}$ , Time to Peak, Maximum Height, Total Flight Time, and Range for a spear thrown at $18 {ms}^{- 1}$ at 30°.

3. Excellence Air Resistance Rationale: 1-paragraph explanation of how real-world air resistance distorts the theoretical parabola (reducing max height, flight time, and shortening range).

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 2 diagram clearly shows v_x remaining constant at all points along the parabolic flight path while v_y is zero at maximum height."

Teacher Planning & NCEA Alignment

NCEA Level 2 Physics Alignment (6 Credits External):

2D Projectile Motion: Demonstrate understanding of horizontal and vertical components of parabolic motion under gravity.
Vector Resolution: Resolve initial velocity vectors using trigonometry ( $v \cos θ, v \sin θ$ ).

Vocabulary: Projectile, trajectory, parabola, horizontal component ( $v_{x}$ ), vertical component ( $v_{y}$ ), gravity ( $g = 9.8 {ms}^{- 2}$ ), peak height, range, flight time.

Same concept, shorter route: the Guided Viewing & Problem Practice version of this unit covers it in Kinematics, Vectors and Projectile Motion. That route is a compact viewing-and-practice sequence; this one builds the concept over ten sections with a formative portfolio.

Lesson 2: Projectile Motion (Horizontal & Angle Vectors)