Problem 69 - Thomas Calculus

Question

Maximum height of a vertically moving body 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 maximum height is \( \frac{v_0^2}{2g} + s_0 \).

Step 1: Identify the Objective

We need to find the maximum height of a body moving vertically, given the height function \( s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \). This requires determining the value of \( t \) at which \( s(t) \) is at its peak.

Step 2: Finding the Time of Maximum Height

The height function \( s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \) is a quadratic function in the standard form \( ax^2 + bx + c \), where \( a = -\frac{1}{2} g \), \( b = v_0 \), and \( c = s_0 \). The maximum point of a parabola that opens downwards (as indicated by the negative leading coefficient \( a \)) occurs at \( t = -\frac{b}{2a} \).

Step 3: Calculate the Time of Maximum Height

Substituting the values for \( a \) and \( b \) into the formula for the vertex, we find the time of maximum height: \[ t = -\frac{v_0}{2 \left(-\frac{1}{2}g\right)} = \frac{v_0}{g} \].

Step 4: Calculate the Maximum Height

Substitute \( t = \frac{v_0}{g} \) back into the original height function: \[ s\left(\frac{v_0}{g}\right) = -\frac{1}{2}g\left( \frac{v_0}{g} \right)^2 + v_0\left(\frac{v_0}{g}\right) + s_0 \].Simplifying yields:\[ s = -\frac{1}{2} \frac{v_0^2}{g} + \frac{v_0^2}{g} + s_0 = \frac{v_0^2}{2g} + s_0 \].

Key Concepts

Quadratic Equations

A quadratic equation is a fundamental expression in algebra, represented by the formula \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations form the shape of a parabola when graphed. In our scenario, the given height function \( s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \), is a specific type of quadratic equation, reflecting the motion of a projectile.

• A quadratic term \( ax^2 \), representing the effect of gravity on the height.
• A linear term \( bx \), showing the initial velocity's effect.
• A constant \( c \), indicating the initial height from the ground level.

Quadratic equations can uniquely predict the behavior of systems exhibiting curved paths, such as those commonly seen in physics with projectile motion. Understanding these equations helps in determining points like maximum or minimum values crucial in applied situations.

Vertex of a Parabola

The vertex of a parabola is the pinnacle point, either the maximum or minimum. For a downward-facing parabola, given by a quadratic equation with a negative \( a \) value, the vertex represents the maximum point.
In the formula \( ax^2 + bx + c \), the vertex \( (h, k) \) of the parabola can be calculated using the formula \( t = -\frac{b}{2a} \). This calculation gives us the time at which the projectile achieves its maximum height.

• The vertex formula highlights the balance point of the curve.
• In our height function, applying the vertex formula finds when the body is at its highest.

For projectile motion, finding the vertex allows us to determine the time and height at the peak, a fundamental step in solving real-world problems.

Projectile Motion

Projectile motion describes the trajectory of an object thrown into space, influenced only by gravity. The motion is inherently two-dimensional, involving both horizontal and vertical components. The vertical motion of a projectile can be modeled by a quadratic equation.

• The initial velocity \( v_0 \) contributes to the upward movement.
• Gravity \( g \) acts downward, affecting the object's height over time.
• The height equation \( s(t) \) helps determine the maximum height reached.

In the height equation \( s = -\frac{1}{2} g t^2 + v_0 t + s_0 \), the motion results from the competing influences of initial velocity and gravitational deceleration. Understanding projectile motion and its equations is crucial for comprehending dynamic systems in both academic and applied physics contexts.

Height Function

The height function, \( s(t) = -\frac{1}{2} g t^2 + v_0 t + s_0 \), represents the vertical position of a projectile over time. By analyzing this function, one can determine how high a projectile flies at any moment and ultimately find the maximum height.

• The term \(-\frac{1}{2} g t^2\) represents the effect of gravity, pulling the object down at an accelerating rate.
• The term \(v_0 t\) captures the initial upward thrust given to the projectile.
• The constant \(s_0\) signifies the starting height position relative to a reference point.

To find the maximum height, calculate \( t = \frac{v_0}{g} \), where the projectile stops rising and begins to fall. Substitute back into the height function to find \( s = \frac{v_0^2}{2g} + s_0 \), revealing the apex of the trajectory. This height function is a keystone of understanding how various forces influence a moving object over time.