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Effect of Gravity on Earth If a rock falls from a height of 20 meters on Earth, the height \(H\) (in meters) after \(x\) seconds is approximately $$ H(x)=20-4.9 x^{2} $$ (a) What is the height of the rock when \(x=1\) second? When \(x=1.1\) seconds? When \(x=1.2\) seconds? (b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters? (c) When does the rock strike the ground?

Short Answer

Expert verified
Heights are 15.1 m, 14.071 m, and 12.944 m at 1, 1.1, and 1.2 seconds. Height is 15 m at 1.01 seconds, 10 m at 1.43 seconds, and 5 m at 1.75 seconds. Rock strikes the ground at 2.02 seconds.

Step by step solution

01

Substitute 1 second into the height formula

To find the height of the rock when the time is 1 second, substitute \(x = 1\) into the formula \(H(x) = 20 - 4.9x^2\): \[ H(1) = 20 - 4.9(1)^2 = 20 - 4.9 = 15.1 \text{ meters} \] So, the height when \(x = 1\) second is 15.1 meters.
02

Substitute 1.1 seconds into the height formula

Next, substitute \(x = 1.1\) into the formula: \[ H(1.1) = 20 - 4.9(1.1)^2 \] Calculate \( (1.1)^2 = 1.21\): \[ H(1.1) = 20 - 4.9 \times 1.21 = 20 - 5.929 = 14.071 \text{ meters} \] So, the height when \(x = 1.1\) seconds is 14.071 meters.
03

Substitute 1.2 seconds into the height formula

Lastly, substitute \(x = 1.2\) into the formula: \[ H(1.2) = 20 - 4.9(1.2)^2 \] Calculate \( (1.2)^2 = 1.44\): \[ H(1.2) = 20 - 4.9 \times 1.44 = 20 - 7.056 = 12.944 \text{ meters} \] So, the height when \(x = 1.2\) seconds is 12.944 meters.
04

Set height to 15 meters and solve for time

To find when the height is 15 meters, set \(H(x) = 15\) and solve: \[ 15 = 20 - 4.9x^2 \] Rearrange and solve for \(x\): \[ 4.9x^2 = 5 \] \[ x^2 = \frac{5}{4.9} \] \[ x = \sqrt{\frac{5}{4.9}} \] \[ x \approx 1.01 \text{ seconds} \]
05

Set height to 10 meters and solve for time

Similarly, for a height of 10 meters, set \(H(x) = 10\) and solve: \[ 10 = 20 - 4.9x^2 \] Rearrange and solve for \(x\): \[ 4.9x^2 = 10 \] \[ x^2 = \frac{10}{4.9} \] \[ x = \sqrt{\frac{10}{4.9}} \] \[ x \approx 1.43 \text{ seconds} \]
06

Set height to 5 meters and solve for time

For a height of 5 meters, set \(H(x) = 5\) and solve: \[ 5 = 20 - 4.9x^2 \] Rearrange and solve for \(x\): \[ 4.9x^2 = 15 \] \[ x^2 = \frac{15}{4.9} \] \[ x = \sqrt{\frac{15}{4.9}} \] \[ x \approx 1.75 \text{ seconds} \]
07

Find when the rock strikes the ground

To find the time when the rock strikes the ground, set \(H(x) = 0\) and solve: \[ 0 = 20 - 4.9x^2 \] Rearrange and solve for \(x\): \[ 4.9x^2 = 20 \] \[ x^2 = \frac{20}{4.9} \] \[ x = \sqrt{\frac{20}{4.9}} \] \[ x \approx 2.02 \text{ seconds} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations are fundamental in various branches of mathematics and physics. Quadratic equations usually appear in the form:\( ax^2 + bx + c = 0 \).

These equations are named quadratic because 'quadra' means square, referring to the term \(x^2\). In our exercise, the height function \(H(x) = 20 - 4.9x^2\) is a quadratic equation with respect to time. Solving quadratic equations can often require methods like factoring, completing the square, or the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

Quadratic equations help model various real-world situations, such as physics problems like the one involving the free fall of a rock. They enable us to calculate the height, time, and other variables related to motion under gravity.
Gravity Effect in Physics
Gravity is one of the fundamental forces of nature, and it affects the motion of all objects with mass. On Earth, the acceleration due to gravity is approximately \(9.8 \text{ m/s}^2\), represented by the constant \(g\). This constant plays a crucial role in free-fall problems.

In free fall, an object is moving under the influence of gravity alone, with no external forces like air resistance affecting its motion. The equations of motion for free-falling objects are derived from the second law of motion by Newton. Specifically, the height function in our exercise, \(H(x) = 20 - 4.9x^2\), is derived from integrating \(a = -g\). The term \(4.9\) comes from \(\frac{1}{2}g\), as it appears in the equations of motion for freely falling objects from a known height. Understanding the effect of gravity helps explain why objects accelerate downwards and how their positions change over time.
Time Calculation
Time calculation is crucial in physics problems, especially those involving motion. To determine how long it takes for an event to occur, such as a rock hitting the ground, you need to solve for time using the relevant equations.

In our exercise, to find the time when the rock strikes the ground, we set the height function to zero and solve for \(x\), as the height at ground level is zero. This approach transforms the problem into solving a quadratic equation. For instance, solving \(0 = 20 - 4.9x^2\) gives us:\[ x = \sqrt{\frac{20}{4.9}} \approx 2.02 \text{ seconds} \].

Besides knowing how to solve quadratic equations, understanding how to manipulate formulas and extract meaningful time values is essential. Concepts like extracting roots, manipulating square terms, and handling various algebraic steps all come into play in these calculations. Mastery of time calculation enables you to determine key motion parameters effectively in a wide range of physics problems.

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Most popular questions from this chapter

True or False If no domain is specified for a function \(f,\) then the domain of \(f\) is the set of real numbers.

Motion of a Golf Ball A golf ball is hit with an initial velocity of 130 feet per second at an inclination of \(45^{\circ}\) to the horizontal. In physics, it is established that the height \(h\) of the golf ball is given by the function $$ h(x)=\frac{-32 x^{2}}{130^{2}}+x $$ where \(x\) is the horizontal distance that the golf ball has traveled. (a) Determine the height of the golf ball after it has traveled 100 feet. (b) What is the height after it has traveled 300 feet? (c) What is \(h(500) ?\) Interpret this value. (d) How far was the golf ball hit? (e) Use a graphing utility to graph the function \(h=h(x)\). (f) Use a graphing utility to determine the distance that the ball has traveled when the height of the ball is 90 feet. (g) Create a TABLE with TblStart \(=0\) and \(\Delta \mathrm{Tbl}=25 .\) To the nearest 25 feet, how far does the ball travel before it reaches a maximum height? What is the maximum height? (h) Adjust the value of \(\Delta\) Tbl until you determine the distance, to within 1 foot, that the ball travels before it reaches its maximum height.

Suppose that the function \(y=f(x)\) is increasing on the interval [-1,5] (a) Over what interval is the graph of \(y=f(x+2)\) increasing? (b) Over what interval is the graph of \(y=f(x-5)\) increasing? (c) Is the graph of \(y=-f(x)\) increasing, decreasing, or neither on the interval [-1,5]\(?\) (d) Is the graph of \(y=f(-x)\) increasing, decreasing, or neither on the interval [-5,1]\(?\)

Answer the questions about each function. $$f(x)=\frac{12 x^{4}}{x^{2}+1}$$ (a) Is the point (-1,6) on the graph of \(f ?\) (b) If \(x=3,\) what is \(f(x) ?\) What point is on the graph of \(f ?\) (c) If \(f(x)=1,\) what is \(x ?\) What point(s) are on the graph of \(f ?\) (d) What is the domain of \(f ?\) (e) List the \(x\) -intercepts, if any, of the graph of \(f\). (f) List the \(y\) -intercept, if there is one, of the graph of \(f\).

Determine algebraically whether each function is even, odd, or neither. \(h(x)=\frac{x}{x^{2}-1}\)

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