A Stone Is Thrown From The Top Of A Tall Cliff. Its Acceleration Is A Constant $-32 \frac{ft}{\text{sec}^2}$ (so $A(t) = -32$). Its Velocity After 2 Seconds Is $18 \frac{ft}{\text{sec}}$, And Its Height After 2 Seconds Is 232

Introduction

When a stone is thrown from the top of a tall cliff, it experiences a constant downward acceleration due to gravity. This acceleration is a fundamental concept in physics, and understanding it is crucial for analyzing the motion of objects under the influence of gravity. In this article, we will delve into the physics behind the motion of the stone, exploring its acceleration, velocity, and height as a function of time.

The Acceleration of the Stone

The acceleration of the stone is a constant $-32 \frac{ft}{\text{sec}^2}$, which is the acceleration due to gravity on Earth. This acceleration is directed downward, and its magnitude is independent of the mass of the stone. The acceleration is given by the equation $A(t) = -32$, where $t$ is the time in seconds.

The Velocity of the Stone

The velocity of the stone after 2 seconds is given as $18 \frac{ft}{\text{sec}}$. To understand how the velocity of the stone changes over time, we need to use the equation of motion for an object under constant acceleration. The equation is given by:

$$v(t) = v_0 + A(t)t$$

where $v(t)$ is the velocity at time $t$, $v_0$ is the initial velocity, and $A(t)$ is the acceleration. Since the acceleration is constant, we can rewrite the equation as:

$$v(t) = v_0 + (-32)t$$

We are given that the velocity after 2 seconds is $18 \frac{ft}{\text{sec}}$, so we can substitute this value into the equation to find the initial velocity:

$$18 = v_0 + (-32)(2)$$

Solving for $v_0$, we get:

$$v_0 = 18 + 64 = 82 \frac{ft}{\text{sec}}$$

The Height of the Stone

The height of the stone after 2 seconds is given as 232 feet. To understand how the height of the stone changes over time, we need to use the equation of motion for an object under constant acceleration. The equation is given by:

$$h(t) = h_0 + v_0t + \frac{1}{2}A(t)t^2$$

where $h(t)$ is the height at time $t$, $h_0$ is the initial height, $v_0$ is the initial velocity, and $A(t)$ is the acceleration. Since the acceleration is constant, we can rewrite the equation as:

$$h(t) = h_0 + v_0t + \frac{1}{2}(-32)t^2$$

We are given that the height after 2 seconds is 232 feet, so we can substitute this value into the equation to find the initial height:

$$232 = h_0 + (82)(2) + \frac{1}{2}(-32)(2)^2$$

Solving for $h_0$, we get:

$$h_0 = 232 - 164 - 64 = 4 \text{ feet}$$

The Motion of the Stone

Now that we have found the initial velocity and height of the stone, we can use the equations of motion to analyze its motion over time. The velocity of the stone as a function of time is given by:

$$v(t) = 82 - 32t$$

The height of the stone as a function of time is given by:

$$h(t) = 4 + 82t - 16t^2$$

We can use these equations to find the velocity and height of the stone at any time $t$.

Conclusion

In this article, we have analyzed the motion of a stone thrown from the top of a tall cliff. We have found the acceleration, velocity, and height of the stone as a function of time, and used these equations to understand its motion over time. The acceleration of the stone is a constant $-32 \frac{ft}{\text{sec}^2}$, and its velocity and height can be found using the equations of motion. This analysis provides a fundamental understanding of the physics behind the motion of objects under the influence of gravity.

References

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

Appendix

Derivation of the Equations of Motion

The equations of motion for an object under constant acceleration are given by:

$$v(t) = v_0 + A(t)t$$

$$h(t) = h_0 + v_0t + \frac{1}{2}A(t)t^2$$

where $v(t)$ is the velocity at time $t$, $v_0$ is the initial velocity, $A(t)$ is the acceleration, $h(t)$ is the height at time $t$, and $h_0$ is the initial height.

To derive these equations, we can use the definition of acceleration as the rate of change of velocity:

$$A(t) = \frac{dv}{dt}$$

Substituting this expression into the equation for velocity, we get:

$$v(t) = v_0 + \int A(t)dt$$

Evaluating the integral, we get:

$$v(t) = v_0 + A(t)t$$

Similarly, we can derive the equation for height by using the definition of acceleration as the rate of change of velocity:

$$A(t) = \frac{dv}{dt}$$

Substituting this expression into the equation for height, we get:

$$h(t) = h_0 + \int v(t)dt$$

Evaluating the integral, we get:

$$h(t) = h_0 + v_0t + \frac{1}{2}A(t)t^2$$

Q&A: Frequently Asked Questions About the Motion of the Stone

Q: What is the acceleration of the stone?

A: The acceleration of the stone is a constant $-32 \frac{ft}{\text{sec}^2}$, which is the acceleration due to gravity on Earth.

Q: What is the velocity of the stone after 2 seconds?

A: The velocity of the stone after 2 seconds is given as $18 \frac{ft}{\text{sec}}$.

Q: What is the height of the stone after 2 seconds?

A: The height of the stone after 2 seconds is given as 232 feet.

Q: How does the velocity of the stone change over time?

A: The velocity of the stone changes over time according to the equation:

$$v(t) = 82 - 32t$$

Q: How does the height of the stone change over time?

A: The height of the stone changes over time according to the equation:

$$h(t) = 4 + 82t - 16t^2$$

Q: What is the initial velocity of the stone?

A: The initial velocity of the stone is 82 $\frac{ft}{\text{sec}}$.

Q: What is the initial height of the stone?

A: The initial height of the stone is 4 feet.

Q: How can I use the equations of motion to analyze the motion of the stone?

A: You can use the equations of motion to analyze the motion of the stone by substituting the given values into the equations and solving for the velocity and height of the stone at any time $t$.

Q: What are the limitations of the equations of motion?

A: The equations of motion are limited to objects that are under the influence of a constant acceleration. If the acceleration is not constant, the equations of motion will not be accurate.

Q: How can I apply the equations of motion to real-world problems?

A: You can apply the equations of motion to real-world problems by substituting the given values into the equations and solving for the velocity and height of the object at any time $t$. This can be useful in a variety of fields, including physics, engineering, and astronomy.

Q: What are some common applications of the equations of motion?

A: Some common applications of the equations of motion include:

• Calculating the trajectory of a projectile
• Analyzing the motion of a pendulum
• Studying the motion of a planet or moon
• Designing a roller coaster or other amusement park ride

Conclusion

In this article, we have answered some frequently asked questions about the motion of a stone thrown from the top of a tall cliff. We have discussed the acceleration, velocity, and height of the stone as a function of time, and used the equations of motion to analyze its motion over time. The equations of motion are a fundamental tool in physics and can be applied to a wide range of real-world problems.

References

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

Appendix

Derivation of the Equations of Motion

The equations of motion for an object under constant acceleration are given by:

$$v(t) = v_0 + A(t)t$$

$$h(t) = h_0 + v_0t + \frac{1}{2}A(t)t^2$$

where $v(t)$ is the velocity at time $t$, $v_0$ is the initial velocity, $A(t)$ is the acceleration, $h(t)$ is the height at time $t$, and $h_0$ is the initial height.

To derive these equations, we can use the definition of acceleration as the rate of change of velocity:

$$A(t) = \frac{dv}{dt}$$

Substituting this expression into the equation for velocity, we get:

$$v(t) = v_0 + \int A(t)dt$$

Evaluating the integral, we get:

$$v(t) = v_0 + A(t)t$$

Similarly, we can derive the equation for height by using the definition of acceleration as the rate of change of velocity:

$$A(t) = \frac{dv}{dt}$$

Substituting this expression into the equation for height, we get:

$$h(t) = h_0 + \int v(t)dt$$

Evaluating the integral, we get:

$$h(t) = h_0 + v_0t + \frac{1}{2}A(t)t^2$$

These equations provide a fundamental understanding of the physics behind the motion of objects under the influence of gravity.