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Mastering Projectile Motion: A Step-by-Step Guide

1. Topic Overview

Projectile motion refers to the motion of an object thrown into the air under the influence of gravity. It is an essential concept in physics, appearing in sports (e.g., basketball), engineering (e.g., rocket launches), and everyday life (e.g., throwing a ball).

2. Key Concepts

• Projectile: Any object moving through space under only the influence of gravity.
• Horizontal Motion: Constant velocity (no acceleration) since gravity doesn’t affect it.
• Vertical Motion: Accelerated motion due to gravity ($9.8 \text{ m/s}^2$ downward).
• Independence of Motion: Horizontal and vertical motions are independent of each other.

Key Equations:

1. Horizontal Motion:
   • $ x = v_{0x} t $
   • $ v_x = v_{0x} $ (constant)
2. Vertical Motion:
   • $ y = v_{0y}t – \frac{1}{2} g t^2 $
   • $ v_y = v_{0y} – g t $
3. Time of Flight (for symmetric motion):
   • $ t = \frac{2 v_{0y}}{g} $
4. Maximum Height:
   • $ h = \frac{v_{0y}^2}{2g} $
5. Range (Horizontal Distance):
   • $ R = \frac{v_0^2 \sin 2\theta}{g} $

3. Step-by-Step Explanation

1. Understand the Motion Components:
   • Motion is divided into horizontal (constant velocity) and vertical (accelerated motion due to gravity).
   • The horizontal component remains unchanged, while the vertical component follows free-fall physics.
2. Find Horizontal and Vertical Components of Initial Velocity:
   • Given initial speed $ v_0 $ and launch angle $ \theta $, split it into:
     • $ v_{0x} = v_0 \cos\theta $
     • $ v_{0y} = v_0 \sin\theta $
3. Analyze Time of Flight:
   • Use vertical motion equations to determine total time in the air.
4. Find Maximum Height:
   • Maximum height is reached when $ v_y = 0 $.
5. Calculate Range:
   • The total horizontal distance covered during flight.

4. Worked Example

Problem: A projectile is launched with a speed of 20 m/s at an angle of 30°. Find:
a) Time of flight
b) Maximum height
c) Range

Step 1: Break Down the Motion

• $ v_{0x} = 20 \cos 30^\circ = 17.32 $ m/s
• $ v_{0y} = 20 \sin 30^\circ = 10 $ m/s

Step 2: Time of Flight

Using $ t = \frac{2 v_{0y}}{g} $:
$ t = \frac{2(10)}{9.8} = 2.04 $ seconds

Step 3: Maximum Height

Using $ h = \frac{v_{0y}^2}{2g} $:
$ h = \frac{10^2}{2(9.8)} = 5.1 $ meters

Step 4: Range

Using $ R = v_{0x} t $:
$ R = (17.32)(2.04) = 35.3 $ meters

5. Practice Problems

1. A ball is thrown at 25 m/s at an angle of 40°. Calculate its:
   • Time of flight
   • Maximum height
   • Range
2. A rock is thrown horizontally from a 45 m high cliff with a speed of 15 m/s.
   • How long does it take to hit the ground?
   • How far does it land from the base of the cliff?

6. Quick Summary & Key Takeaways

✅ Projectile motion is a combination of horizontal motion (constant velocity) and vertical motion (accelerated by gravity).
✅ Use trigonometry to break velocity into horizontal and vertical components.
✅ The horizontal motion does not change, while vertical motion follows free-fall physics.
✅ Time of flight depends on vertical motion, and range depends on horizontal motion.

7. Self-Check Quiz

Q1: What happens to the horizontal velocity of a projectile during flight?
Q2: What equation would you use to find the range of a projectile?
Q3: If two projectiles are launched at complementary angles (e.g., 30° and 60°), which will travel farther?

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