A Rocket Is Launched From A Tower. The Height Of The Rocket, $y$ In Feet, Is Related To The Time After Launch, $x$ In Seconds, By The Given Equation. Using This Equation, Find The Time That The Rocket Will Hit The Ground, To The

Introduction

Understanding the Problem The height of a rocket, $y$ in feet, is related to the time after launch, $x$ in seconds, by a given equation. In this problem, we are tasked with finding the time that the rocket will hit the ground. To do this, we need to use the given equation to determine the time at which the height of the rocket is zero, indicating that it has reached the ground.

The Equation of Motion

The equation of motion for the rocket is given by:

$$y = -16x^2 + 128x$$

where $y$ is the height of the rocket in feet and $x$ is the time after launch in seconds.

Finding the Time of Impact

To find the time of impact, we need to set the height of the rocket, $y$, to zero and solve for $x$. This is because when the rocket hits the ground, its height is zero.

Setting $y = 0$, we get:

$$0 = -16x^2 + 128x$$

Solving the Quadratic Equation

The equation above is a quadratic equation in the form of $ax^2 + bx + c = 0$, where $a = -16$, $b = 128$, and $c = 0$. We can solve this equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Substituting the values of $a$, $b$, and $c$ into the quadratic formula, we get:

$$x = \frac{-128 \pm \sqrt{128^2 - 4(-16)(0)}}{2(-16)}$$

Simplifying the Expression

Simplifying the expression under the square root, we get:

$$x = \frac{-128 \pm \sqrt{16384}}{-32}$$

$$x = \frac{-128 \pm 128}{-32}$$

Finding the Two Possible Solutions

There are two possible solutions to the equation:

$$x = \frac{-128 + 128}{-32} = \frac{0}{-32} = 0$$

$$x = \frac{-128 - 128}{-32} = \frac{-256}{-32} = 8$$

Interpreting the Results

The two possible solutions to the equation are $x = 0$ and $x = 8$. However, the solution $x = 0$ is not physically meaningful, as it represents the time at which the rocket was launched, not the time at which it hit the ground. Therefore, the time at which the rocket will hit the ground is $x = 8$ seconds.

Conclusion

In this problem, we used the given equation of motion to find the time at which the rocket will hit the ground. By setting the height of the rocket to zero and solving the resulting quadratic equation, we found that the time of impact is $x = 8$ seconds.

Discussion

The equation of motion for the rocket is a quadratic equation in the form of $ax^2 + bx + c = 0$. The solution to this equation is given by the quadratic formula, which involves the square root of the discriminant. In this case, the discriminant is $b^2 - 4ac$, which is equal to $16384$. The square root of this value is $128$, which is a positive number. Therefore, the two possible solutions to the equation are $x = \frac{-128 + 128}{-32} = 0$ and $x = \frac{-128 - 128}{-32} = 8$. The solution $x = 0$ is not physically meaningful, as it represents the time at which the rocket was launched, not the time at which it hit the ground. Therefore, the time at which the rocket will hit the ground is $x = 8$ seconds.

Final Answer

The final answer is $\boxed{8}$ seconds.

Introduction

In our previous article, we explored the problem of finding the time that a rocket will hit the ground after being launched from a tower. We used the given equation of motion to determine the time of impact, and found that the rocket will hit the ground in 8 seconds.

Q&A

Q: What is the equation of motion for the rocket?

A: The equation of motion for the rocket is given by:

$$y = -16x^2 + 128x$$

where $y$ is the height of the rocket in feet and $x$ is the time after launch in seconds.

Q: How do we find the time of impact?

A: To find the time of impact, we need to set the height of the rocket, $y$, to zero and solve for $x$. This is because when the rocket hits the ground, its height is zero.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations of the form $ax^2 + bx + c = 0$. The formula is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Q: How do we use the quadratic formula to solve the equation of motion?

A: We substitute the values of $a$, $b$, and $c$ into the quadratic formula, and then simplify the expression under the square root. This gives us the two possible solutions to the equation.

Q: What are the two possible solutions to the equation?

A: The two possible solutions to the equation are $x = 0$ and $x = 8$. However, the solution $x = 0$ is not physically meaningful, as it represents the time at which the rocket was launched, not the time at which it hit the ground.

Q: What is the final answer?

A: The final answer is $\boxed{8}$ seconds.

Additional Questions and Answers

Q: What is the significance of the discriminant in the quadratic formula?

A: The discriminant is the expression under the square root in the quadratic formula. It is given by $b^2 - 4ac$. If the discriminant is positive, then the quadratic equation has two distinct solutions. If the discriminant is zero, then the quadratic equation has one repeated solution. If the discriminant is negative, then the quadratic equation has no real solutions.

Q: How do we interpret the results of the quadratic formula?

A: The results of the quadratic formula give us the two possible solutions to the equation. We need to check which solution is physically meaningful, and discard the other solution.

Q: What are some common applications of the quadratic formula?

A: The quadratic formula has many applications in physics, engineering, and other fields. Some common applications include finding the time of impact of a projectile, determining the stability of a system, and solving optimization problems.

Conclusion

In this article, we have explored the problem of finding the time that a rocket will hit the ground after being launched from a tower. We used the given equation of motion to determine the time of impact, and found that the rocket will hit the ground in 8 seconds. We also answered some common questions about the quadratic formula and its applications.

Final Answer

The final answer is $\boxed{8}$ seconds.