Derivation of the Equation of Motion

Oct 13, 2025 by recoveryshake.com 74 views

Introduction

When a ball is thrown vertically into the air, it follows a curved trajectory under the influence of gravity. The motion of the ball can be described using the equations of motion, which relate the position, velocity, and acceleration of the ball at any given time. In this article, we will explore the physics behind the motion of a ball thrown vertically into the air, and derive the equation of motion for the height of the ball as a function of time.

The Equation of Motion

The height of the ball at time $t$ (in seconds) is given by the function $h(t) = 43t - 4.9t^2 \, \text{m}$ . This equation describes the motion of the ball under the influence of gravity, where the initial velocity is $43 \, \text{m/s}$ and the acceleration due to gravity is $-4.9 \, \text{m/s}^2$ .

To derive this equation, we can use the following kinematic equations:

$v = u + at$
$s = ut + \frac{1}{2}at^2$

where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time, and $s$ is the displacement.

Derivation of the Equation of Motion

Let's assume that the ball is thrown from ground level with an initial velocity of $43 \, \text{m/s}$ . We can use the kinematic equations to derive the equation of motion for the height of the ball.

First, we can use the equation $v = u + at$ to find the final velocity of the ball at time $t$ . Since the acceleration due to gravity is $-4.9 \, \text{m/s}^2$ , the final velocity will be:

$v = 43 - 4.9t$

Next, we can use the equation $s = ut + \frac{1}{2}at^2$ to find the displacement of the ball at time $t$ . Since the initial velocity is $43 \, \text{m/s}$ , the displacement will be:

$s = 43t - 4.9t^2$

However, this equation describes the displacement of the ball, not its height. To find the height of the ball, we need to subtract the displacement from the initial height (which is zero, since the ball is thrown from ground level).

Therefore, the equation of motion for the height of the ball is:

$h(t) = 43t - 4.9t^2$

Interpretation of the Equation

The equation $h(t) = 43t - 4.9t^2$ describes the motion of the ball under the influence of gravity. The first term, $43t$ , represents the initial velocity of the ball, which is $43 \, \text{m/s}$ . The second term, $-4.9t^2$ , represents the acceleration due to gravity, which is $-4.9 \, \text{m/s}^2$ .

As the ball rises, the height of the ball will increase due to the initial velocity, but will decrease due to the acceleration due to gravity. At the peak of the motion, the height of the ball will be maximum, and will then decrease as the ball falls back to the ground.

Graphical Representation

To visualize the motion of the ball, we can plot the equation $h(t) = 43t - 4.9t^2$ as a function of time.

import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0, 10, 100)
h = 43*t - 4.9*t**2

plt.plot(t, h)
plt.xlabel('Time (s)')
plt.ylabel('Height (m)')
plt.title('Motion of the Ball')
plt.grid(True)
plt.show()

This code will generate a plot of the height of the ball as a function of time, which will help us visualize the motion of the ball.

Conclusion

In this article, we have derived the equation of motion for the height of a ball thrown vertically into the air. The equation $h(t) = 43t - 4.9t^2$ describes the motion of the ball under the influence of gravity, where the initial velocity is $43 \, \text{m/s}$ and the acceleration due to gravity is $-4.9 \, \text{m/s}^2$ . We have also interpreted the equation and provided a graphical representation of the motion of the ball.

Future Work

In future work, we can use this equation to analyze the motion of the ball under different initial conditions, such as different initial velocities or angles of projection. We can also use this equation to design experiments to measure the acceleration due to gravity, or to study the effects of air resistance on the motion of the ball.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Note: The references provided are for general physics textbooks and are not specific to the topic of the ball thrown vertically into the air.

Introduction

In our previous article, we explored the physics behind the motion of a ball thrown vertically into the air. We derived the equation of motion for the height of the ball as a function of time, and interpreted the results. In this article, we will answer some common questions related to the motion of the ball.

Q: What is the initial velocity of the ball?

A: The initial velocity of the ball is $43 \, \text{m/s}$ .

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is $-4.9 \, \text{m/s}^2$ .

Q: What is the equation of motion for the height of the ball?

A: The equation of motion for the height of the ball is $h(t) = 43t - 4.9t^2$ .

Q: What is the maximum height of the ball?

A: To find the maximum height of the ball, we need to find the time at which the height is maximum. We can do this by taking the derivative of the equation of motion with respect to time, and setting it equal to zero.

\frac{dh}{dt} = 43 - 9.8t = 0

Solving for $t$ , we get:

t = \frac{43}{9.8} = 4.39 \, \text{s}

Substituting this value of $t$ into the equation of motion, we get:

h(4.39) = 43(4.39) - 4.9(4.39)^2 = 19.1 \, \text{m}

Therefore, the maximum height of the ball is $19.1 \, \text{m}$ .

Q: How long does it take for the ball to reach the ground?

A: To find the time it takes for the ball to reach the ground, we need to find the time at which the height is zero. We can do this by setting the equation of motion equal to zero and solving for $t$ .

h(t) = 43t - 4.9t^2 = 0

Solving for $t$ , we get:

t = \frac{43}{4.9} = 8.78 \, \text{s}

Therefore, it takes $8.78 \, \text{s}$ for the ball to reach the ground.

Q: What is the velocity of the ball at the maximum height?

A: To find the velocity of the ball at the maximum height, we need to take the derivative of the equation of motion with respect to time, and evaluate it at the time at which the height is maximum.

\frac{dh}{dt} = 43 - 9.8t

Evaluating this expression at $t = 4.39 \, \text{s}$ , we get:

\frac{dh}{dt} = 43 - 9.8(4.39) = 0

Therefore, the velocity of the ball at the maximum height is zero.

Q: What is the velocity of the ball when it reaches the ground?

A: To find the velocity of the ball when it reaches the ground, we need to take the derivative of the equation of motion with respect to time, and evaluate it at the time at which the height is zero.

\frac{dh}{dt} = 43 - 9.8t

Evaluating this expression at $t = 8.78 \, \text{s}$ , we get:

\frac{dh}{dt} = 43 - 9.8(8.78) = -43 \, \text{m/s}

Therefore, the velocity of the ball when it reaches the ground is $-43 \, \text{m/s}$ .

Conclusion

In this article, we have answered some common questions related to the motion of a ball thrown vertically into the air. We have used the equation of motion to find the maximum height, the time it takes for the ball to reach the ground, the velocity of the ball at the maximum height, and the velocity of the ball when it reaches the ground.

Future Work

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.