Projectile Motion Companion Notes
Physics Reference Guide
Projectile Motion Companion Notes
Student Guide for Packet: U02L08 - Projectile Motion Intro
Name:
Date:
1. Do Now: Free Fall Secret
Mass Does NOT Affect Fall Time! In free fall (negligible air resistance), all dropped objects accelerate at \( g = 10\text{ m/s}^2 \).
• \( t = \sqrt{2h/g} \) → Since \(h=10\text{m}\), all rocks take same time.
→ Packet Answer: Choose "All the same"
2. Predictions: What is a Projectile?
Any airborne object whose flight is governed solely by gravity after launch. No pushing force remains in flight!
• Examples: Tossed paper wad, thrown ball, launched cannonball.
• Key Factors: Launch speed (\(v_0\)), angle (\(\theta\)), height (\(h\)), gravity (\(g\)).
The Golden Law: Horizontal & Vertical Independence
Core Physics Principle
Horizontal and vertical motions occur simultaneously but do not affect one another. They only share air time (\(t\)).
Horizontal (X) Motion \(a_x = 0\)
- • Net Force: \( F_{\text{net}, x} = 0\text{ N} \) (no horizontal push or drag)
- • Acceleration: \( a_x = 0\text{ m/s}^2 \) (never speeds up/slows down)
- • Velocity: \( v_x = \text{constant} \) everywhere in flight
- • Distance (Range): \( d_x = v_{0x} \cdot t \)
Vertical (Y) Motion \(a_y = -10\text{ m/s}^2\)
- • Net Force: \( F_{\text{net}, y} = F_g = mg \) (acts DOWNWARD)
- • Acceleration: \( a_y = -10\text{ m/s}^2 \) (constant downward gravity)
- • Velocity: Decreases rising, \(0\) at peak, increases falling
- • Peak Fact: At top peak, \( v_y = 0\text{ m/s} \), but \( a_y = 10\text{ m/s}^2 \) down!
Free Body Diagram Rule (Apply Findings #13 & Exit Ticket #1)
Common Trap!
The Rule: Once in the air, the thrower's hand is gone! The ONLY force acting at ANY point is gravity (\(F_g\)) pointing straight down.
Point A (Rising)
\(F_g\)
Only gravity downward
Point B (Apex / Peak)
\(F_g\)
Still only gravity downward!
Point C (Falling)
\(F_g\)
Only gravity downward!
Free Fall, Angled-Launched (ALP), and Horizontally-Launched (HLP): ALL FBDs HAVE ONLY \(F_g\) DOWN!
Exit Ticket Acceleration Solution Table
Standard convention: \( g = 10\text{ m/s}^2 \) down
| Object Position | Horizontal Accel. (\(a_x\)) | Vertical Accel. (\(a_y\)) | Why? (Newton's 2nd Law) |
|---|
| Point A (Rising / Launch) | 0 m/s² | 10 m/s² down | \(F_x = 0 \implies a_x = 0\); Gravity acts down |
| Point B (Peak / Apex) | 0 m/s² | 10 m/s² down | Velocity \(v_y = 0\), but gravity NEVER turns off! |
| Point C (Falling / Landing) | 0 m/s² | 10 m/s² down | Gravity continues accelerating mass down |
Physics U02L08: Projectile Motion Intro • Conceptual Foundation Page 1 of 2
Part 2: Trajectory & Calculations
Launch Angles & Homework 2.08 Master Guide
Packet Reference: Figure 1 & Table 1
Key Patterns from Table 1 & Figure 1
1 Maximum Range: 45°
At any given launch speed, 45° produces the maximum range (\(163\text{ m}\) at \(40\text{ m/s}\)).
Optimal balance between hang time (\(t\)) & forward speed (\(v_x\)).
2 Complementary Pairs
Any two angles that sum to 90° yield the exact same range on level ground!
• 15° & 75° → 82 m
• 30° & 60° → 141 m
• 40° & 50° → 161 m
3 Peak Height & Time
As angle increases toward 90°, vertical speed (\(v_{0y}\)) increases.
• Higher angle = Higher peak (\(h_{\text{max}}\))
• Higher angle = Longer hang time (\(t_{\text{total}}\))
• Max peak is at 90° (straight up).
Step-by-Step Problem Solving (HW 2.08)
Packet Pages 7 & 8
Question 1: Largest Range?
Look at Table 1 for Range values. The highest range listed is 163 meters at 45°.
Answer: (C) 163 meters
Question 2: Highest Peak?
Height depends on vertical velocity component (\(v_{0y}\)). Largest angle gives highest peak.
Answer: (D) Launch angle = 85°
Question 3: Angle vs Range Rule?
15° & 75° both equal 82m; 30° & 60° both equal 141m. Angles that add to 90° yield equal range.
Answer: (B) Angles adding to 90°
Question 4: Golfer Clearing Tree?
Tree is 30m tall at 50m out. At \(x=50\text{ m}\), Figure 1 shows 45° trajectory reaches ~35m and maximizes distance.
Answer: (C) 45 degrees
Question 5: Isolating \(v_{0y}\) Variable?
To test if vertical velocity causes hang time changes, hold \(v_{0x}\) constant (at 0) by firing straight up!
Answer: (D) Launch straight up (90°)
Question 6: Cannon Over 300m?
Fall time \( t = \sqrt{2(22)/10} = 2.1\text{ s} \). Range = \( 140 \times 2.1 = 294\text{ m} \). To reach >300m: increase speed (\(v_{xi} > 143\text{ m/s}\)) or increase cliff height.
Essential Formulas
• Range (Horizontal): \( d_x = v_{0x} \cdot t \)
• Fall Time (Cliff/Drop): \( t = \sqrt{\frac{2h}{g}} \)
• Total Hang Time: \( t_{\text{total}} = \frac{2v_{0y}}{g} \)
• Max Peak Height: \( h_{\text{max}} = \frac{v_{0y}^2}{2g} \)
Avoid These 3 Common Traps!
❌ "Forward velocity slows down in air."
✓ No! \(v_x\) is constant because \(F_x = 0\).
❌ "At peak, acceleration is 0."
✓ No! \(v_y = 0\), but \(a_y = 10\text{ m/s}^2\) downward.
❌ "Mass affects fall rate."
✓ No! All objects accelerate at \(g\).
Physics U02L08: Projectile Motion Intro • Trajectory & Problem Solving Page 2 of 2
