Question

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

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

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.

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.

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.

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 \(x\): \[ 4.9x^2 = 5 \] \[ x^2 = \frac{5}{4.9} \] \[ x = \sqrt{\frac{5}{4.9}} \] \[ x \approx 1.01 \text{ seconds} \]

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 \(x\): \[ 4.9x^2 = 10 \] \[ x^2 = \frac{10}{4.9} \] \[ x = \sqrt{\frac{10}{4.9}} \] \[ x \approx 1.43 \text{ seconds} \]

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 \(x\): \[ 4.9x^2 = 15 \] \[ x^2 = \frac{15}{4.9} \] \[ x = \sqrt{\frac{15}{4.9}} \] \[ x \approx 1.75 \text{ seconds} \]

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 \(x\): \[ 4.9x^2 = 20 \] \[ x^2 = \frac{20}{4.9} \] \[ x = \sqrt{\frac{20}{4.9}} \] \[ x \approx 2.02 \text{ seconds} \]

Key concepts

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.