Problem 28 - Differential Equations and Linear Algebra

Question

An object that is initially thrown vertically upward with a speed of 2 meters/second from a height of \(h\) meters takes 10 seconds to reach the ground. Set up and solve the initial-value problem that governs the motion of the object, and determine \(h\)

Step-by-Step Solution

Answer

The object was initially at a height of \(h = 470\ \mathrm{m}\) above the ground and the initial-value problem that governs its motion is \(h = v_0t + \frac{1}{2}at^2\), where \(v_0 = 2\ \mathrm{m/s}\), \(a = -9.8\ \mathrm{m/s^2}\), and \(t = 10\ \mathrm{s}\).

Step 1: Write down the known values and symbols

We know the following values: - Initial velocity, \(v_0=2 \mathrm{m/s}\) - Time taken to reach the ground, \(t=10 \mathrm{s}\) - Acceleration due to gravity, \(a=-g=-9.8 \mathrm{m/s^2}\) (The negative sign indicates that gravity acts in the downward direction) We need to find out the height, \(h\).

Step 2: Apply the equation of motion

We use the equation of motion \(s = v_0t + \frac{1}{2}at^2\), where \(s\) is the displacement, \(v_0\) is the initial velocity, \(t\) is time, and \(a\) is acceleration: \(h = v_0t + \frac{1}{2}at^2\)

Step 3: Substitute the known values

Now, let's substitute the known values into the equation: \(h = (2\ \mathrm{m/s})(10\ \mathrm{s}) + \frac{1}{2}(-9.8\ \mathrm{m/s^2})(10\ \mathrm{s})^2\)

Step 4: Simplify and solve for h

Simplify the equation to find the value of \(h\): \(h = 20\ \mathrm{m} - 490\ \mathrm{m}\) \(h = -470\ \mathrm{m}\) Since the height can never be negative, we need to understand what the negative sign indicates: The object was thrown upward and took 10 seconds to come back to the ground, so it must have moved \(470\) meters downward from the initial height in that time. Therefore, the object was initially at a height of \(h = 470\ \mathrm{m}\) above the ground.

Key Concepts

Initial-Value Problem

An initial-value problem in differential equations consists of finding a function that satisfies a given differential equation and initial conditions. In the context of this exercise, the problem involves determining the height from which an object is thrown upward based on its motion over a specific time. Here, the initial condition provides essential details about the motion at the beginning:

• Initial velocity: 2 meters/second.
• Time taken to reach the ground: 10 seconds.

The given initial conditions are used to derive a solution that describes the entire system's behavior at all times. The solution of this initial-value problem involves applying the laws of physics, specifically, using kinematic equations to link the motion parameters in a coherent manner. This helps accurately compute variables such as height or displacement.

Equation of Motion

The equation of motion is crucial for predicting the dynamics of a moving object under the influence of forces. In the context of the given problem, we employ one of the basic kinematic equations:\[s = v_0t + \frac{1}{2}at^2\]where:

• \(s\) represents the displacement of the object relative to its starting point, which is our unknown height.
• \(v_0\) is the initial velocity, given as 2 meters/second.
• \(t\) is the duration, 10 seconds, during which the object is in motion.
• \(a\) stands for the acceleration, which in this case is the force due to gravity, acting downward and represented as \(-9.8\) meters/second squared.

This equation helps link initial velocity, acceleration, and time to determine the displacement, thereby revealing insights about the object's starting point.

Gravity

Gravity is a universal force that attracts objects towards one another, and on Earth, it gives weight to physical objects and causes them to fall toward the ground. In our problem, gravity is the sole force acting on the object after it is thrown. Its effect is characterized by an acceleration of approximately \(9.8 \mathrm{m/s^2}\), pulling the object downward. The negative sign in the equation \(a = -9.8 \mathrm{m/s^2}\) highlights the direction of gravity relative to the initial upward motion. It indicates that while the object was initially projected upward, gravity counteracts this motion by pulling it back toward the ground.As the object ascends, gravity gradually slows it down until it reaches the highest point. After that, gravity accelerates its descent until it reaches the surface. Understanding gravity's role in this scenario helps accurately calculate various motion parameters, including the height from which the object was launched.