Lesson 5.1: Modelling Projectile Motion

Introduction

In this lesson, we will explore the fascinating world of projectile motion. Understanding how objects move through the air is essential for various fields, including physics, engineering, and sports. We will model the motion of a projectile as a particle moving freely under the influence of gravity in a vertical plane. By the end of this lesson, students will be able to:

• Model a projectile as a particle moving freely under gravity.
• Understand the independence of horizontal and vertical motion.
• Resolve an initial velocity into horizontal and vertical components.
• State the modelling assumptions for projectile motion.
• Calculate the timing and distance related to projectile motion.

The Nature of Projectile Motion

Independent Motion Components

When analyzing projectile motion, it is crucial to understand that the horizontal and vertical components of the motion are independent of each other. This independence allows us to simplify the analysis of projectile trajectories.

1. Horizontal Motion:

The horizontal motion of the projectile is characterized by a constant velocity. This is because, in the absence of air resistance, there are no horizontal forces acting on the projectile. Therefore, the horizontal component of motion can be described using the equation:

$$x = ut \quad (1)$$

where:

• $ x $ is the horizontal displacement,
• $ u $ is the initial horizontal velocity, and
• $ t $ is the time of flight.

1. Vertical Motion:

In contrast, the vertical motion is affected by the force of gravity, resulting in a constant acceleration downwards. The vertical motion can be described by the following equations

$y = ut + \frac{1}{2}gt^2 \quad (2)$

$v = u + gt \quad (3)$

where:

• $ y $ is the vertical displacement,
• $ v $ is the final vertical velocity,
• $ g $ is the acceleration due to gravity $(g = -9.8 \, \text{m/s}^2)$, and
• $ u $ is the initial vertical velocity.

Worked Example

Consider a ball thrown with an initial speed $ u = 20 \, \text{m/s} $ at an angle $ \theta = 30^\circ $ from the horizontal. We need to resolve $ u $ into its horizontal ($ u_x $) and vertical ($ u_y $) components.

• First, let's use trigonometric functions:

$$u_x = u \cdot \cos(\theta) \quad (4)$$

$$u_y = u \cdot \sin(\theta) \quad (5)$$

Now substituting the values:

$$u_x = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32 \, \text{m/s}$$

$$u_y = 20 \cdot \sin(30^\circ) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$$

The initial velocity can thus be resolved into:

• Horizontal component: $ u_x \approx 17.32 \, \text{m/s} $
• Vertical component: $ u_y = 10 \, \text{m/s} $

Modelling Assumptions

When dealing with projectile motion, several assumptions are made to simplify calculations:

1. The object is treated as a particle: This means we neglect the size and shape of the object for the purpose of calculating the motion.
2. Air resistance is negligible: In many problems, we ignore the effects of air resistance to simplify calculations.
3. The only force acting on the object is gravity: Gravity acts downwards with a constant acceleration of approximately $ 9.8 \, \text{m/s}^2 $.
4. The velocity of projection is instantaneously at an angle: The initial velocity is considered to act at a specific angle to the horizontal.

These assumptions are crucial for applying the equations of motion effectively.

Time of Flight

The time of flight $ T $ can be determined by examining the vertical motion of the projectile. The projectile will return to the original level of launch when the vertical displacement $ y = 0 $. Using equation (2) above:

$$0 = u_y \cdot T + \frac{1}{2}(-g)T^2 \quad (6)$$

This simplifies to:

$$0 = 10T - 4.9T^2$$

Factoring out $ T $:

$$T(10 - 4.9T) = 0$$

Setting this equal to zero leads to two possible solutions:

1. $ T = 0 $ (the time at the launch, which is not useful to us)
2. $ T = \frac{10}{4.9} \approx 2.04 \, \text{s} $

Thus, the time of flight is approximately $ 2.04 \, \text{s} $.

Range of the Projectile

The range $ R $ of the projectile is the horizontal distance it travels during the time of flight. Using equation (1) for horizontal motion:

$$R = u_x \cdot T \quad (7)$$

Substituting the values found earlier:

$$R \approx 17.32 \cdot 2.04 \approx 35.34 \, \text{m}$$

Thus, the range of the projectile is approximately $ 35.34 \, \text{m} $.

Maximum Height

Another key aspect of projectile motion is the maximum height $ H $. This occurs at the peak of the vertical component of the trajectory. To find the maximum height, we can use the following kinematic equation:

$$H = \frac{u_y^2}{2g} \quad (8)$$

Substituting the values:

$$H = \frac{10^2}{2 \cdot 9.8} \approx \frac{100}{19.6} \approx 5.10 \, \text{m}$$

Therefore, the maximum height achieved by the projectile is approximately $ 5.10 \, \text{m} $.

Conclusion

In this lesson, students learned the principles of modeling projectile motion, including the independence of horizontal and vertical motion, resolving velocities, and key results like time of flight, range, and maximum height. Understanding these concepts lays the foundation for analyzing more complex motions and real-world applications.

Study Notes

• Projectile motion is modeled as a particle under gravity.
• Horizontal motion has constant velocity; vertical motion has constant acceleration.
• Resolve initial velocity into horizontal and vertical components using trigonometric functions.
• The acceleration due to gravity is $ g = 9.8 \, \text{m/s}^2 $.
• Time of flight can be calculated by considering vertical motion and setting $ y = 0 $.
• The range can be calculated using horizontal motion equations, and maximum height is derived from the vertical motion equations.