Problem 6 - Algebra 2

Question

The height \(h\) of an object \(t\) seconds after it is dropped is given by \(h=-\frac{1}{2} g t^{2}+h_{0},\) where \(h_{0}\) is the initial height and \(g\) is the acceleration due to gravity. The acceleration due to gravity near Earth's surface is \(9.8 \mathrm{m} / \mathrm{s}^{2},\) while on Jupiter it is 23.1 \(\mathrm{m} / \mathrm{s}^{2} .\) Suppose an object is dropped from an initial height of 100 meters from the surface of each planet. Find the time it takes for the object to reach the ground on each planet to the nearest tenth of a second.

Step-by-Step Solution

Answer

4.5 seconds on Earth, 2.9 seconds on Jupiter.

Step 1: Understand the Formula

The height of the object over time is given by the formula \(h = -\frac{1}{2} g t^2 + h_0\). Our goal is to find the time \(t\) when \(h = 0\) (when the object hits the ground).

Step 2: Solve for Earth

For Earth, we use \(g = 9.8\, \text{m/s}^2\) and \(h_0 = 100\, \text{m}\). Set \(h = 0\) and solve: \(0 = -\frac{1}{2} \times 9.8 \times t^2 + 100\). This simplifies to: \(4.9t^2 = 100\).

Step 3: Calculate Time for Earth

Rearrange equation to solve for \(t\): divide both sides by 4.9, resulting in \(t^2 = \frac{100}{4.9}\). Compute \(t^2 \approx 20.41\). Taking the square root gives \(t \approx \sqrt{20.41} \) which is approximately 4.5 seconds.

Step 4: Solve for Jupiter

For Jupiter, use \(g = 23.1\, \text{m/s}^2\) and \(h_0 = 100\, \text{m}\). Again, set \(h = 0\) and solve: \(0 = -\frac{1}{2} \times 23.1 \times t^2 + 100\). Simplify to: \(11.55t^2 = 100\).

Step 5: Calculate Time for Jupiter

Rearrange to solve for \(t\): divide both sides by 11.55, so \(t^2 = \frac{100}{11.55}\). Compute \(t^2 \approx 8.66\). Taking the square root gives \(t \approx \sqrt{8.66}\), which is approximately 2.9 seconds.

Key Concepts

Quadratic Equation

A quadratic equation is a mathematical expression that takes the form \( ax^2 + bx + c = 0 \). It includes terms with different powers of the variable.
In the context of projectile motion, the equation gives us the relationship between time and height.

• The standard form is useful because it allows us to find the point at which the object hits the ground, that is when the height \( h = 0 \).
• By setting the equation \( h = 0 \), we transform it into a quadratic equation, making \( t^2 \) the subject.

To solve for \( t \), we rearrange the equation and apply algebraic operations like squaring to find when the object reaches ground level.

Gravity

Gravity is the force of attraction by a planet that pulls objects towards its center. It is represented by \( g \) and varies amongst planets.

• On Earth, gravity is \( 9.8\, \text{m/s}^2 \), while on Jupiter, it is stronger at \( 23.1\, \text{m/s}^2 \).
• The greater the gravitational pull, the faster the object will fall.

This concept is crucial when calculating the time taken to reach the ground in projectile motion problems, similar to the one given. Understanding how differences in gravity affect motion helps in accurately predicting motion on various planets.

Kinematics

Kinematics deals with the motion of objects, describing its path based solely on initial velocity and acceleration without considering the forces involved.

• This topic explores how an object's speed and height change over time under gravitational influence.
• The kinematic equation \( h = -\frac{1}{2}gt^2 + h_0 \) used here specifically models an object moving under the influence of gravity.

Breaking down kinematics helps students understand how gravity impacts acceleration and height. The initial height \( h_0 \) and gravitational pull \( g \) are key in calculating time \( t \), which is the variable representing how long it takes for an object to hit the ground.

Problem Solving

Problem-solving involves several steps in mathematics and physics. Breaking problems into manageable steps can ease understanding and solution processes.

• First, interpret the problem and identify known variables: \( g \), \( h_0 \), and the desired \( h \).
• Second, use the appropriate formula, here derived from the kinematic equations.
• Convert the expression into a quadratic equation by setting \( h = 0 \).

By systematically solving, you find \( t \) for different values of \( g \). Practice in problem-solving can increase confidence in tackling more complex kinematic problems.