Lesson 5.2: Time of Flight and Maximum Height

Introduction

In this lesson, we will explore the concepts of time of flight and maximum height for projectiles. Understanding these concepts is crucial for solving problems involving two-dimensional motion under the influence of gravity. By the end of this lesson, students will be able to:

• Use the vertical component of motion to find the time to maximum height and the total time of flight.
• Calculate the greatest height reached by a projectile.
• Determine the velocity and direction of motion of a projectile at a given time.
• Find the time of flight of a projectile launched from and returning to a horizontal plane.
• Calculate the maximum height reached using the vertical component of the initial velocity.

Hook

Imagine you are at a sports stadium, and you throw a ball straight up into the air. How long do you think it will take for that ball to reach its highest point? What about the total time the ball will be in the air before it comes back down? In this lesson, we will answer these questions and more as we delve into the fascinating world of projectile motion.

Understanding Projectile Motion

Before we dive into calculations, let's review what we mean by projectile motion. A projectile is any object that is thrown into the air and moves under the influence of gravity. In our study of projectiles, we will consider the following:

1. The motion in the vertical direction is influenced by gravity, which acts downwards.
2. The horizontal motion is uniform, meaning there is no acceleration in that direction (ignoring air resistance).

This leads us to treat the horizontal and vertical motions as independent of each other. This independence allows us to analyze projectile motion using kinematic equations.

Time to Maximum Height

How to Calculate Time to Maximum Height

The time taken for a projectile to reach its maximum height can be determined using the vertical component of the initial velocity. This component can be found using:

$$v_{y0} = v_0 \cdot \sin(\theta)$$

where:

• $v_0$ is the initial velocity of the projectile,
• $\theta$ is the angle of projection,
• $v_{y0}$ is the initial vertical velocity component.

At the maximum height, the vertical velocity becomes zero ($v_y = 0$). We can use the following kinematic equation to find the time taken to reach this height:

$$v_y = v_{y0} - g \cdot t$$

Setting $v_y = 0$ gives:

$$0 = v_{y0} - g \cdot t_{max}$$

Rearranging gives:

$$t_{max} = \frac{v_{y0}}{g}$$

Worked Example 1: Finding Time to Maximum Height

Let's say we launch a projectile with an initial velocity of $30$ m/s at an angle of $45^\circ$. We first find the vertical component of the initial velocity:

$$v_{y0} = v_0 \cdot \sin(45^\circ) = 30 \cdot \frac{\sqrt{2}}{2} \approx 21.21 \text{ m/s}$$

Now, we can determine the time to reach maximum height:

$$t_{max} = \frac{v_{y0}}{g} = \frac{21.21}{9.81} \approx 2.16 \text{ s}$$

Thus, it will take approximately $2.16$ seconds for the projectile to reach its maximum height.

Total Time of Flight

How to Calculate Total Time of Flight

The total time of flight can be found by realizing that the time taken to go up to the maximum height is equal to the time taken to come back down. Therefore, the total time of flight ($T$) is:

$$T = 2 \cdot t_{max}$$

Worked Example 2: Finding Total Time of Flight

Using the previous example, we can find the total time of flight:

$$T = 2 \cdot t_{max} = 2 \cdot 2.16 \approx 4.32 \text{ s}$$

So, the total time the projectile is in the air is approximately $4.32$ seconds.

Maximum Height

How to Calculate Maximum Height

To find the maximum height ($H$) reached by the projectile, we can use the vertical motion equations. The formula for maximum height is:

$$H = v_{y0} \cdot t_{max} - \frac{1}{2} g \cdot (t_{max})^2$$

Worked Example 3: Finding Maximum Height

Continuing from our previous examples, we can calculate the maximum height:

First, we know:

• $v_{y0} \approx 21.21 \text{ m/s}$
• $t_{max} \approx 2.16 \text{ s}$

Now, substituting into the height equation:

$$H = 21.21 \cdot 2.16 - \frac{1}{2} \cdot 9.81 \cdot (2.16)^2$$

Calculating this gives:

$$H \approx 45.8 \text{ m} - 21.5 \text{ m} \approx 24.3 \text{ m}$$

Thus, the projectile reaches a maximum height of approximately $24.3$ meters.

Velocity and Direction of Motion

How to Calculate Velocity at a Given Time

The velocity of a projectile at any time $t$ can be calculated by finding both the horizontal and vertical components.

The horizontal component of velocity is constant and given by:

$$v_x = v_0 \cdot \cos(\theta)$$

The vertical component is influenced by gravity, given by:

$$v_y = v_{y0} - g \cdot t$$

Thus, the overall velocity can be represented as a vector combining these two components.

Worked Example 4: Finding Velocity at a Specific Time

Assuming $t = 1$ second into the previous example:

1. Horizontal velocity:

$$v_x = 30 \cdot \cos(45^\circ) \approx 21.21 \text{ m/s}$$

1. Vertical velocity:

$$v_y = 21.21 - 9.81 \cdot 1 \approx 11.4 \text{ m/s}$$

Now, the total velocity ($v$) can be found using:

$$v = \sqrt{v_x^2 + v_y^2} \approx \sqrt{(21.21)^2 + (11.4)^2} \approx 24.32 \text{ m/s}$$

We can find the direction using the arctangent function:

$$\theta = \tan^{-1}\left(\frac{v_y}{v_x} ight) \approx \tan^{-1}\left(\frac{11.4}{21.21} ight) \approx 28.7^\circ$$

Thus, at $t = 1$ second, the projectile’s velocity is approximately $24.32$ m/s at an angle of $28.7^\circ$ above the horizontal.

Conclusion

In this lesson, we have learned how to calculate the time to maximum height, total time of flight, maximum height reached, and the velocity and direction of motion of a projectile at a given time. By understanding these concepts, students can solve various problems related to projectile motion and apply this knowledge in real-world situations like sports, engineering, and physics research.

Study Notes

• The vertical motion of a projectile is influenced by gravity, leading to a downward acceleration of $9.81 \text{ m/s}^2$.
• Time to maximum height can be calculated using $t_{max} = \frac{v_{y0}}{g}$ where $v_{y0}$ is the vertical component of initial velocity.
• The total time of flight is $T = 2 \cdot t_{max}$.
• Maximum height can be calculated using $H = v_{y0} \cdot t_{max} - \frac{1}{2} g \cdot (t_{max})^2$.
• The horizontal component of velocity remains constant during projectile motion, while the vertical component changes due to gravity.