Problem 36 - Advanced Engineering Mathematics

Question

Suppose a small cannonball weighing \(16 \mathrm{lb}\) is shot vertically upward with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). The answer to the question, "How high does the cannonball go?" depends on whether we take air resistance into account. (a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by \(d^{2} s / d t^{2}=-g\) (equation (12) of Section 1.3). Since \(d s / d t=v(t)\) the last differential equation is the same as \(d v / d t=-g\), where we take \(g=32 \mathrm{ft} / \mathrm{s}^{2} .\) Find the velocity \(v(t)\) of the cannonball at time \(t\). (b) Use the result obtained in part (a) to determine the height \(s(t)\) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.

Step-by-Step Solution

Answer

The maximum height is 1406.25 feet.

Step 1: Understanding the Equation of Motion

The problem states that the acceleration due to gravity is constant, thus we have \( \frac{dv}{dt} = -g \), where \( g = 32 \ \text{ft/s}^2 \). This equation models the motion of the cannonball under the influence of gravity alone, ignoring air resistance.

Step 2: Integrating to Find Velocity

To find the velocity function \( v(t) \), we integrate the equation \( \frac{dv}{dt} = -g \). This integration gives us \( v(t) = -gt + C \), where \( C \) is the constant of integration. Given that the initial velocity \( v(0) = 300 \ \text{ft/s} \), we substitute to find \( C \): \[ 300 = -g \cdot 0 + C \Rightarrow C = 300 \]. Therefore, \( v(t) = -32t + 300 \).

Step 3: Integrating Velocity to Find Position

Now, use the velocity function to find the position function \( s(t) \). Since \( \frac{ds}{dt} = v(t) \), integrate \( v(t) = -32t + 300 \) to find \( s(t) \). This results in \( s(t) = \int (-32t + 300) \, dt = -16t^2 + 300t + D \), where \( D \) is a constant of integration. Assuming the cannonball is fired from ground level, \( s(0) = 0 \), we find \( D \). Substituting \( t = 0 \), we get \[ 0 = -16(0)^2 + 300(0) + D \Rightarrow D = 0 \]. Thus, \( s(t) = -16t^2 + 300t \).

Step 4: Determine the Maximum Height of the Cannonball

The maximum height is attained when the velocity is zero \( v(t) = 0 \). From \( v(t) = -32t + 300 \): \[ 0 = -32t + 300 \Rightarrow t = \frac{300}{32} = 9.375 \ \text{seconds} \]. Substitute \( t = 9.375 \ \text{s} \) into \( s(t) \) to find the maximum height: \[ s(9.375) = -16(9.375)^2 + 300(9.375) = -16(87.890625) + 2812.5 = 1406.25 \ \text{feet} \].

Key Concepts

Differential Equations

Differential equations are mathematical expressions involving functions and their derivatives. They are essential in modeling the behavior of dynamic systems, especially in physics where they describe things like motion, growth, or decay. In our exercise, the key differential equation is \( \frac{dv}{dt} = -g \). This equation tells us that the rate at which the velocity \( v(t) \) of the cannonball changes over time \( t \) is equal to the negative of the gravitational force \( g \), given as \( 32 \ \text{ft/s}^2 \).

• The negative sign in the equation indicates that gravity is acting in the opposite direction of the cannonball's initial upward motion.
• Solving this differential equation by integrating gives us the velocity equation; it displays how the velocity of the cannonball decreases over time due to gravity.

By understanding and working with these equations, we can predict and analyze the future state of the projectile, such as its velocity at any given time.

Kinematics

Kinematics deals with the motion of objects without considering the forces causing the motion. It involves parameters like displacement, velocity, and acceleration. In our situation, a cannonball is shot upwards, and we need to determine how it moves over time.
Initially, we know:

• The initial velocity \( v_0 = 300 \ \text{ft/s} \)
• Acceleration due to gravity \( g = 32 \ \text{ft/s}^2 \)

Through integration, we derive that the velocity function \( v(t) = -32t + 300 \). This equation tells us the cannonball slows down as it rises due to gravity.
To find the position or height \( s(t) \), we integrate the velocity equation to get \( s(t) = -16t^2 + 300t \). This function represents how the height changes with time, showing a curve that ascends and then descends, visually representing the cannonball's path.

Maximum Height Calculation

To find how high the cannonball goes, we look for the point where its upward speed is zero—the maximum height. This occurs when \( v(t) = 0 \). We set \( -32t + 300 = 0 \) and solve for \( t \):
\[ t = \frac{300}{32} = 9.375 \ \text{seconds} \]
This time indicates when the cannonball reaches its peak height.
Substitute \( t = 9.375 \ \text{s} \) into the height function \( s(t) = -16t^2 + 300t \):
\[ s(9.375) = -16(9.375)^2 + 300(9.375) \]
Calculating this gives us:
\[ s(9.375) = -16(87.890625) + 2812.5 = 1406.25 \ \text{feet} \]
Thus, the cannonball reaches a maximum height of 1406.25 feet. This illustrates the beautiful arc where the cannonball gently slows, pauses momentarily at its peak, and then begins its descent due to gravity.