Lesson 2.3: Vertical Motion Under Gravity

Introduction

Vertical motion under gravity is a fundamental concept in kinematics, describing how objects move when subject to the acceleration due to gravity, denoted as $ g $. For most purposes, we consider $ g = 9.8 \, \text{m/s}^2 $ as a constant near the Earth's surface. This lesson will cover how to model free vertical motion, determine maximum heights, and calculate times of flight for objects projected vertically. By the end of this lesson, you will be able to apply the SUVAT equations to vertical motion effectively.

Learning Objectives

• Model free vertical motion as constant acceleration with $ g = 9.8 \, \text{m/s}^2 $
• Choose positive directions and analyze objects thrown upward and falling back down
• Calculate time to greatest height, maximum height achieved, and total time of flight
• Apply SUVAT formulae with $ a = g $ (or $ a = -g $)
• Determine the maximum height and total time of flight for a vertically projected particle

Vertical Motion Basics

When dealing with vertical motion, it's important to establish a frame of reference. Typically, we choose upwards as the positive direction, but downwards can also be chosen based on context. Understanding this choice is crucial since it affects the signs of our acceleration and displacement.

Setting Up the Motion

1. Choosing a Positive Direction:

• Upward is positive: Displacement and velocity upwards are positive, while downward displacement and velocity are negative.
• Downward is positive: The opposite sign convention, where everything upwards is negative and downward is positive, can also be used but is less common.

1. Acceleration Due to Gravity:

The acceleration of any object in free fall near the Earth's surface is directed downwards and has a magnitude of $ g = 9.8 \, \text{m/s}^2 $. If we choose upwards as positive, then:

• $ a = -g = -9.8 \, \text{m/s}^2 $

Worked Example 1: Free Fall Calculation

Problem: An object is dropped from a height of 20 meters.

1. Determine how long it takes to reach the ground and the impact velocity just before it hits the ground.

Solution:

• Initial height ($ s_0 $): 20 m
• Final height ($ s $): 0 m (ground)
• Initial velocity ($ u $): 0 m/s (dropped)
• Acceleration ($ a $): $ -9.8 \, \text{m/s}^2 $

Using the SUVAT equation, we can find the time taken to hit the ground:

$$s = ut + \frac{1}{2} a t^2$$

Substituting the known values:

$$0 = 20 + 0 \cdot t + \frac{1}{2} (-9.8) t^2$$

This simplifies to:

$$0 = 20 - 4.9t^2$$

Thus,

$$4.9t^2 = 20$$

$$t^2 = \frac{20}{4.9} \approx 4.08$$

$$t \approx \sqrt{4.08} \approx 2.02\, \text{s}$$

• Next, to find the impact velocity, we use another SUVAT equation:

$$v = u + at$$

$$v = 0 + (-9.8)(2.02) \approx -19.8 \, \text{m/s}$$

The negative sign indicates the direction is downwards.

Thus, the object takes approximately $ 2.02 \, \text{s} $ to hit the ground and has an impact velocity of approximately $ -19.8 \, \text{m/s} $.

Vertical Projection

Now let's consider vertical motion where an object is projected upwards. In such cases, the initial velocity plays a crucial role in determining the time to reach the maximum height and its total flight time.

Key Variables in Vertical Projection

• Initial velocity ($ u $): The speed at which the object is projected upwards.
• Final velocity ($ v $): At the peak height, the velocity is $ 0 \, \text{m/s} $.
• Acceleration ($ a $): If we assume the upward direction is positive, then $ a = -9.8 \, \text{m/s}^2 $. This means the velocity decreases as the object rises and eventually becomes zero at the maximum height.
• Displacement ($ s $): The maximum height reached above the initial projection point.

Worked Example 2: Maximum Height and Total Time of Flight

Problem: An object is thrown vertically upwards with an initial velocity of $ 20 \, \text{m/s} $. Calculate the maximum height reached and the total time of flight.

Solution:

• Initial velocity ($ u $): $ 20 \, \text{m/s} $
• Final velocity ($ v $): $ 0 \, \text{m/s} $
• Acceleration ($ a $): $ -9.8 \, \text{m/s}^2 $

1. Maximum Height:

Using the SUVAT equation:

$$v^2 = u^2 + 2as$$

Substituting the values we have:

$$0 = (20)^2 + 2(-9.8)s$$

This leads to:

$$0 = 400 - 19.6s$$

$$19.6s = 400$$

$$s = \frac{400}{19.6} \approx 20.41 \, \text{m}$$

The maximum height reached is approximately $ 20.41 \, \text{m} $.

1. Time to Reach Maximum Height:

Using the SUVAT equation:

$$v = u + at$$

We want to find $ t $:

$$0 = 20 + (-9.8)t$$

$$9.8t = 20$$

$$t \approx \frac{20}{9.8} \approx 2.04 \, \text{s}$$

1. Time of Flight:

The total time of flight is double the time taken to reach maximum height because the time taken to ascend is equal to the time taken to descend:

$$\text{Total time} = 2t \approx 2 \times 2.04 \approx 4.08 \, \text{s}$$

Thus, the maximum height reached is approximately $ 20.41 \, \text{m} $ and the total time of flight is approximately $ 4.08 \, \text{s} $.

Conclusion

In this lesson, we explored vertical motion under the influence of gravity, focusing on free fall and vertically projected motion. We established a framework for choosing positive direction conventions, derived the SUVAT equations for vertical acceleration, and worked through both free fall and vertical projection examples. Understanding these principles is essential for analyzing motion in two dimensions and solving complex kinematics problems.

Study Notes

• Objects in free fall accelerate at $ g = 9.8 \, \text{m/s}^2 $ downwards.
• Choose positive direction (up or down) based on the problem context.
• Use the SUVAT equations for calculating time, displacement, and velocity in vertical motion:
• $ v = u + at $
• $ s = ut + \frac{1}{2} at^2 $
• $ v^2 = u^2 + 2as $
• For upward projects, calculate maximum height and time of flight by determining initial velocity and using $ g $.
• The time to reach maximum height equals the time to descend from it back to the original position.