Blog

 

 

If an object is launched with an initial velocity $v_0$ at an angle $\theta$:

• Horizontal velocity: $v_{0x} = v_0 \cos \theta$
• Vertical velocity: $v_{0y} = v_0 \sin \theta$

---

Equations of Projectile Motion

Using kinematic equations, we analyze projectile motion:

1. Time of flight (total time the projectile is in the air):

   $$t = \frac{2 v_{0y}}{g}$$

2. Maximum height (highest point reached):

   $$H = \frac{v_{0y}^2}{2g}$$

3. Range (total horizontal distance covered):

   $$R = \frac{v_0^2 \sin 2\theta}{g}$$

---

Real-World Applications of Projectile Motion

Projectile motion has numerous applications in science and engineering:

1. Sports and Athletics

• Basketball: When shooting a basket, the ball follows a parabolic trajectory.
• Football (Soccer & American Football): Kicking a ball at an angle requires knowledge of optimal angles for maximum distance.

 

2. Engineering and Military

• Rocket Launches: Engineers calculate projectile motion to ensure a rocket reaches orbit.
• Artillery and Missiles: Military experts use projectile motion to determine the trajectory of missiles and bombs.

 

3. Amusement Parks and Roller Coasters

• Roller coasters use projectile motion principles in curved sections and drop sequences.

 

4. Traffic Accidents and Forensics

• In accident investigations, forensic scientists calculate how far a car will travel if it drives off a bridge.

---

Common Mistakes and How to Avoid Them

Common Mistake	Correction
Ignoring horizontal motion	Remember that horizontal velocity remains constant.
Using the same velocity for x and y directions	Always break velocity into components using trigonometry.
Forgetting that time is the same for both motions	The time of flight applies to both horizontal and vertical motion.

---

AP Physics Exam Questions with Solutions

Question 1: A Football Kick

A football is kicked at 20 m/s at an angle of 30°. Find:
a) The time of flight
b) The maximum height
c) The horizontal range

Solution:

Given:

• $v_0 = 20$ m/s
• $\theta = 30^\circ$
• $g = 9.8$ m/s²

Step 1: Find $v_{0x}$ and $v_{0y}$

$$v_{0x}=v_0\cos 30=20\times 0.866=17.32\text{ m/s}$$

$$v_{0x} = v_0 \cos 30 = 20 \times 0.866 = 17.32 \text{ m/s}$$ $$v_{0y} = v_0 \sin 30 = 20 \times 0.5 = 10 \text{ m/s}$$

Step 2: Time of Flight

$$t = \frac{2 v_{0y}}{g} = \frac{2 \times 10}{9.8} = 2.04 \text{ s}$$

Step 3: Maximum Height

$$H = \frac{v_{0y}^2}{2g} = \frac{10^2}{2 \times 9.8} = 5.1 \text{ m}$$

Step 4: Range

$$R = \frac{v_0^2 \sin 2\theta}{g} = \frac{20^2 \sin 60}{9.8} = 35.36 \text{ m}$$

---

Question 2: A Cliff Problem

A stone is thrown horizontally at 15 m/s from a 45 m high cliff. Find:
a) The time taken to hit the ground
b) The horizontal distance traveled

Solution:

Given:

• $v_{0x} = 15$ m/s
• $y = 45$ m
• $g = 9.8$ m/s²

Step 1: Time of Flight
Using $y = \frac{1}{2} g t^2$:

$$45=\frac{1}{2}(9.8)t^2$$

$$45 = \frac{1}{2} (9.8) t^2$$ $$t^2=\frac{90}{9.8}=9.18$$

$$t^2 = \frac{90}{9.8} = 9.18$$ $$t = 3.03 \text{ s}$$

Step 2: Horizontal Distance

$$x = v_{0x} \times t = 15 \times 3.03 = 45.45 \text{ m}$$

---

Projectile motion is a vital topic in physics, used in various real-life applications. If you’re preparing for the AP Physics exam, keep practicing problems and revisiting concepts.

For more practice questions and in-depth explanations, visit:

• www.refresh.com
• blog.refreshkid.com

For additional physics resources, check out the College Board AP Physics Course.

Understanding the basics of motion in two dimensions will help you excel in physics and apply these concepts in real-world situations.