Lesson 5.4: Projectiles from a Height and on Inclined Ground

Introduction

In this lesson, we will explore the fascinating world of projectiles, specifically focusing on projectiles launched from a height above the landing level, such as from a cliff or a table. We will also investigate the special case of horizontal projection, where there is zero initial vertical velocity. Our exploration will equip students with the skills to find where a projectile lands when the launch and landing heights differ. By the end of this lesson, you will understand how to set up the vertical equation for a projectile launched from a height and solve for the time of flight and range when launch and landing levels differ.

Learning Objectives:

• Understand the concept of projectiles launched from a height above the landing level.
• Recognize horizontal projection as a special case with zero initial vertical velocity.
• Learn to find where a projectile lands when the launch and landing heights differ.
• Set up the vertical equation for a projectile launched from a height.
• Solve for the time of flight and range when the launch and landing levels differ.

Understanding Projectiles Launched from a Height

When dealing with projectiles, it's essential to remember that they are influenced by two forces: gravity acting downwards and the initial velocity at which they are launched. In particular, when a projectile is launched from a height, we need to consider how both components of motion—the vertical and horizontal—play a role in the projectile's trajectory.

Key Concepts

1. Vertical and Horizontal Components: When a projectile is launched, we can break its initial velocity into two orthogonal components: horizontal ($v_{0x}$) and vertical ($v_{0y}$). The horizontal component remains constant while the vertical component is influenced by gravity. The equations of motion we will use are:

• Horizontal motion: $ x(t) = v_{0x} t $
• Vertical motion: $ y(t) = h + v_{0y} t - \frac{1}{2} g t^2 $

where $h$ is the initial height from which the projectile is launched, $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$), and $t$ is time.

1. Time of Flight: The total time that the projectile remains in the air depends on the height from which it is launched and the initial vertical velocity.

1. Range: This is the horizontal distance the projectile travels while it's in motion.

Worked Example 1: Projectile Launched from a Height

Let's consider a scenario where a projectile is launched from a height of $h = 20 \, \text{m}$ with an initial velocity of $v_0 = 30 \, \text{m/s}$ at an angle of $45^\circ$ to the horizontal. We need to find the time of flight and the range of the projectile.

Step 1: Break down the initial velocity into components.

The horizontal and vertical components of the velocity are given by:

• Horizontal component: $ v_{0x} = v_0 \cos(45^\circ) = 30 \cdot \frac{\sqrt{2}}{2} = 15\sqrt{2} \, \text{m/s} $
• Vertical component: $ v_{0y} = v_0 \sin(45^\circ) = 30 \cdot \frac{\sqrt{2}}{2} = 15\sqrt{2} \, \text{m/s} $

Step 2: Set up the vertical equation.

$ y(t) = h + v_{0y} t - \frac{1}{2} g t^2 $

We can set $y(t) = 0$ to find the time when the projectile hits the ground:

$ 0 = 20 + 15\sqrt{2} t - \frac{1}{2}(9.81)t^2 $

This gives us:

$ 0 = -4.905t^2 + 15\sqrt{2}t + 20 $

Solving this quadratic equation using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

• Here, $a = -4.905$, $b = 15\sqrt{2}$, and $c = 20$.

Step 3: Calculate Discriminant and Find $t$:

• Discriminant:

$ D = b^2 - 4ac = (15\sqrt{2})^2 - 4(-4.905)(20) $

$ D = 450 - (-392.4) = 842.4 \approx 29.04 $ (approx.)

• Hence,

$ t = \frac{-15\sqrt{2} \pm \sqrt{D}}{2(-4.905)} $

Calculating these with approximations:

• Using $15\sqrt{2} \approx 21.2132$:

$ t = \frac{-21.2132 \pm 29.04}{-9.81} $

We will take only the positive value:

$ t \approx \frac{7.8268}{9.81} \approx 0.797 \text{ seconds} $

Step 4: Find the Range:

Using this time in the horizontal motion equation:

$ x(t) = v_{0x} t = 15\sqrt{2} (0.797) \approx 21.2132 (0.797) \approx 16.89 \text{ meters} $

Common Misconceptions

• Misconception 1: Students often think the horizontal and vertical components affect each other. They do not; horizontal motion remains constant while vertical motion is influenced by gravity.
• Misconception 2: Assuming the time of flight only depends on horizontal distance; it actually depends on both vertical height and vertical velocity.

Conclusion

In this lesson, we have examined the principles behind projectiles launched from a height and the calculations involved in finding the time of flight and range when the launch and landing heights differ. By applying these principles, students can analyze a variety of projectile motion scenarios and deepen their understanding of two-dimensional kinematics.

Study Notes

• Projectiles have independent horizontal and vertical motions.
• Vertical motion is influenced by gravity, while horizontal motion is constant.
• Break initial velocity into horizontal and vertical components.
• Use kinematic equations for vertical motion: $y(t) = h + v_{0y} t - \frac{1}{2} g t^2$.
• To find where a projectile lands, set the vertical motion equation to zero and solve for time.
• Calculate range using $x(t) = v_{0x} t$.
• Watch for misconceptions regarding horizontal and vertical independence.