The Height, { H $}$, Of A Falling Object { T $}$ Seconds After It Is Dropped From A Platform 300 Feet Above The Ground Is Modeled By The Function { H(t) = 300 - 16t^2 $}$.Which Expression Could Be Used To Determine The

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Introduction

When an object is dropped from a certain height, its height above the ground can be modeled using a quadratic function. In this article, we will explore the mathematical model that describes the height of a falling object as a function of time. We will analyze the given function and determine which expression could be used to find the time it takes for the object to hit the ground.

The Mathematical Model

The height, { h $}$, of a falling object { t $}$ seconds after it is dropped from a platform 300 feet above the ground is modeled by the function { h(t) = 300 - 16t^2 $}$. This function represents the height of the object at any given time { t $}$.

Understanding the Function

The given function is a quadratic function, which means it has a parabolic shape. The general form of a quadratic function is { f(x) = ax^2 + bx + c $}$, where { a $}$, { b $}$, and { c $}$ are constants. In this case, the function is { h(t) = 300 - 16t^2 $}$, where { a = -16 $}$, { b = 0 $}$, and { c = 300 $}$.

Finding the Time it Takes for the Object to Hit the Ground

To find the time it takes for the object to hit the ground, we need to find the value of { t $}$ when the height { h(t) $}$ is equal to 0. This is because when the object hits the ground, its height above the ground is 0.

We can set up an equation using the given function:

{ 0 = 300 - 16t^2 $}$

To solve for { t $}$, we can rearrange the equation:

{ 16t^2 = 300 $}$

Dividing both sides by 16:

{ t^2 = \frac{300}{16} $}$

Taking the square root of both sides:

{ t = \pm \sqrt{\frac{300}{16}} $}$

Since time cannot be negative, we take the positive square root:

{ t = \sqrt{\frac{300}{16}} $}$

Simplifying the expression:

{ t = \sqrt{\frac{75}{4}} $}$

{ t = \frac{\sqrt{75}}{2} $}$

{ t = \frac{5\sqrt{3}}{2} $}$

Therefore, the expression that could be used to determine the time it takes for the object to hit the ground is { t = \frac{5\sqrt{3}}{2} $}$.

Conclusion

In this article, we analyzed the mathematical model that describes the height of a falling object as a function of time. We used the given function to find the time it takes for the object to hit the ground. The expression { t = \frac{5\sqrt{3}}{2} $}$ can be used to determine the time it takes for the object to hit the ground.

References

  • [1] "Quadratic Functions". Math Open Reference. Retrieved 2023-02-20.
  • [2] "Parabolas". Math Is Fun. Retrieved 2023-02-20.

Further Reading

  • "The Mathematics of Falling Objects". Mathworld. Retrieved 2023-02-20.
  • "Quadratic Equations". Khan Academy. Retrieved 2023-02-20.
    The Height of a Falling Object: Q&A =====================================

Introduction

In our previous article, we explored the mathematical model that describes the height of a falling object as a function of time. We analyzed the given function and determined which expression could be used to find the time it takes for the object to hit the ground. In this article, we will answer some frequently asked questions related to the height of a falling object.

Q: What is the initial height of the object?

A: The initial height of the object is 300 feet above the ground.

Q: What is the equation of the function that models the height of the object?

A: The equation of the function is { h(t) = 300 - 16t^2 $}$, where { h(t) $}$ is the height of the object at time { t $}$.

Q: What is the value of { t $}$ when the object hits the ground?

A: To find the value of { t $}$ when the object hits the ground, we need to set the height { h(t) $}$ equal to 0 and solve for { t $}$. The expression that could be used to determine the time it takes for the object to hit the ground is { t = \frac{5\sqrt{3}}{2} $}$.

Q: What is the significance of the coefficient { -16 $}$ in the function?

A: The coefficient { -16 $}$ represents the acceleration due to gravity, which is approximately 32 feet per second squared. This means that the object is accelerating downward at a rate of 32 feet per second squared.

Q: How does the function change as time increases?

A: As time increases, the height of the object decreases. This is because the object is accelerating downward due to gravity, causing its height to decrease over time.

Q: Can the function be used to model the height of an object that is thrown upward?

A: No, the function { h(t) = 300 - 16t^2 $}$ is only applicable to objects that are dropped from a certain height. If an object is thrown upward, the function would need to be modified to account for the initial velocity of the object.

Q: How can the function be used in real-world applications?

A: The function { h(t) = 300 - 16t^2 $}$ can be used to model the height of objects in various real-world applications, such as:

  • Calculating the time it takes for a parachute to deploy
  • Determining the height of a building or a bridge
  • Modeling the trajectory of a projectile

Conclusion

In this article, we answered some frequently asked questions related to the height of a falling object. We hope that this Q&A article has provided a better understanding of the mathematical model that describes the height of a falling object as a function of time.

References

  • [1] "Quadratic Functions". Math Open Reference. Retrieved 2023-02-20.
  • [2] "Parabolas". Math Is Fun. Retrieved 2023-02-20.

Further Reading

  • "The Mathematics of Falling Objects". Mathworld. Retrieved 2023-02-20.
  • "Quadratic Equations". Khan Academy. Retrieved 2023-02-20.