Problem 75 - Thomas Calculus in SI Units

Question

The height of a body moving vertically is given by $$ s=-\frac{1}{2} g t^{2}+v_{0} t+s_{0}, \quad g > 0 $$ with \(s\) in meters and \(t\) in seconds. Find the body's maximum height.

Step-by-Step Solution

Answer

The body's maximum height is \( \frac{1}{2}\frac{v_0^2}{g} + s_0 \).

Step 1: Identify the Given Equation

The problem gives the vertical motion equation: \[ s = -\frac{1}{2} g t^2 + v_0 t + s_0 \]where \( s \) is the height in meters, \( t \) is the time in seconds, \( g \) is the acceleration due to gravity, \( v_0 \) is the initial velocity, and \( s_0 \) is the initial height.

Step 2: Determine the Type of Motion

Notice that the equation is a quadratic in the form of \[ s = at^2 + bt + c \]with \[ a = -\frac{1}{2}g,\ b = v_0,\ c = s_0. \]Since \( a < 0 \), it is a downward-opening parabola, indicating there is a maximum height.

Step 3: Calculate the Time of Maximum Height

The vertex of the parabola (i.e., the point of maximum height) is found using \[ t = -\frac{b}{2a} \]Substituting \( b = v_0 \) and \( a = -\frac{1}{2}g \):\[ t = -\frac{v_0}{2(-\frac{1}{2}g)} = \frac{v_0}{g} \]

Step 4: Substitute Time into the Height Equation

To find the maximum height, substitute \( t = \frac{v_0}{g} \) back into the height equation:\[ s = -\frac{1}{2}g\left(\frac{v_0}{g}\right)^2 + v_0\left(\frac{v_0}{g}\right) + s_0 \]Simplify the expression to get:\[ s = -\frac{1}{2}\frac{v_0^2}{g} + \frac{v_0^2}{g} + s_0 \]

Step 5: Simplify the Maximum Height Expression

Combine terms in the expression for maximum height:\[ s = \left(-\frac{1}{2} + 1\right)\frac{v_0^2}{g} + s_0 \]\[ s = \frac{1}{2}\frac{v_0^2}{g} + s_0 \]This expression gives the body's maximum height.

Key Concepts

Maximum Height

In the context of quadratic motion, determining the maximum height of a vertically moving body is a crucial aspect. This maximum height is the highest point the object reaches during its motion. People often encounter this in projectile motion problems where a ball is thrown up, an arrow is shot, or any other object moves in a parabolic path.

To calculate the maximum height, we utilize the given height equation:

• \[ s = -\frac{1}{2}g t^2 + v_0 t + s_0 \]

The maximum height is found by first identifying the time \( t \) at which this height is achieved. We use the vertex formula for this, as explained further below. By substituting this time back into the height equation, we are able to solve for \( s \), the maximum height.

For instance, knowing \( v_0 \), the initial velocity of the object, \( g \), the gravitational acceleration, and \( s_0 \), the initial position, allows us to plug these values into the height equation. By simplifying the expression, we obtain:

• \[ s = \frac{1}{2}\frac{v_0^2}{g} + s_0 \]

This straightforward formula gives us the maximum height reached by the object. Understanding how to derive it ensures one can approach similar problems with confidence and accuracy.

Parabolic Trajectory

The path taken by an object in quadratic motion often resembles a parabola. This is why we refer to it as a parabolic trajectory. Such trajectories are common in physics, especially when analyzing freely falling bodies or projectiles.

• A parabolic trajectory is characterized by the shape of its path, which curves downward.
• This happens due to the gravitational pull acting on the object.
• Gravity causes the object to accelerate downwards, forming a symmetric trajectory about its highest point.

Because of this shape, the object has two main points of interest:

• The point where it starts rising after its launch.
• The apex, or maximum height, it reaches before descending.

Each of these stages of motion can be explained using quadratic equations. As mentioned earlier, the coefficient \( a \) in the equation \( s = at^2 + bt + c \) informs us that the parabola opens downward because \( a < 0 \). This indicates that the object's trajectory has a vertex as the highest point on its path. Understanding the properties of this parabolic path is crucial for solving problems related to motion dynamics.

Vertex Formula

The vertex of a parabola is essential in determining the peak point of a quadratic function. When dealing with the vertical motion, the vertex gives us the time at which an object reaches its maximum height when moving along a parabolic trajectory.

For a quadratic equation \( y = ax^2 + bx + c \), the vertex can be found using the formula:

• \[ t = -\frac{b}{2a} \]

This formula is pivotal in finding the vertex of the parabola representing our motion equation. By substituting \( a = -\frac{1}{2}g \) and \( b = v_0 \), the time at which maximum height occurs is given by:

• \[ t = \frac{v_0}{g} \]

It's crucial to remember that this time doesn't give us the height directly. Instead, it marks the point on the time axis at which the height is maximum. With "\( t \)" determined, you plug it into the height equation to find the precise maximum height. This use of the vertex formula effectively simplifies the problem-solving process, making it a powerful tool for tackling quadratic motion challenges.