Chapter 4: Problem 75
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.
Short Answer
Expert verified
The body's maximum height is \( \frac{1}{2}\frac{v_0^2}{g} + s_0 \).
Step by step solution
01
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.
02
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.
03
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 \( a = -\frac{1}{2}g \):\[ t = -\frac{v_0}{2(-\frac{1}{2}g)} = \frac{v_0}{g} \]
04
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:\[ s = -\frac{1}{2}\frac{v_0^2}{g} + \frac{v_0^2}{g} + s_0 \]
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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:
To calculate the maximum height, we utilize the given height equation:
- \[ s = -\frac{1}{2}g t^2 + v_0 t + s_0 \]
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:
- \[ s = \frac{1}{2}\frac{v_0^2}{g} + s_0 \]
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.
- The point where it starts rising after its launch.
- The apex, or maximum height, it reaches before descending.
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:
For a quadratic equation \( y = ax^2 + bx + c \), the vertex can be found using the formula:
- \[ t = -\frac{b}{2a} \]
- \[ t = \frac{v_0}{g} \]