Problem 36 - Advanced Engineering Mathematics

Question

How High? — No Air Resistance 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 attained by the cannonball is 1406.25 feet.

Step 1: Define the Differential Equation for Velocity

The velocity of the cannonball is given by the differential equation \( \frac{dv}{dt} = -g \), where \( g = 32 \, \text{ft/s}^2 \). This indicates that the velocity decreases linearly over time due to gravity.

Step 2: Integrate to Find Velocity Function

Integrate the differential equation from Step 1: \( \int \frac{dv}{dt} \, dt = \int -g \, dt \). The result is \( v(t) = -gt + C \), where \( C \) is the integration constant.

Step 3: Apply Initial Condition for Velocity

At time \( t = 0 \), the initial velocity \( v(0) = 300 \, \text{ft/s} \) is given. Use this to find the constant \( C \): \( 300 = -32 \times 0 + C \). Thus, \( C = 300 \). So, \( v(t) = -32t + 300 \).

Step 4: Define the Differential Equation for Height

The height function is obtained from the velocity function by integration: \( \frac{ds}{dt} = v(t) = -32t + 300 \).

Step 5: Integrate to Find Height Function

Integrate the equation \( \int \frac{ds}{dt} \, dt = \int (-32t + 300) \, dt \). This yields \( s(t) = -16t^2 + 300t + C' \), where \( C' \) is another integration constant.

Step 6: Apply Initial Condition for Height

At \( t = 0 \), the initial height \( s(0) = 0 \). Substitute to find \( C' \): \( 0 = -16 \times 0^2 + 300 \times 0 + C' \). Thus, \( C' = 0 \). This gives \( s(t) = -16t^2 + 300t \).

Step 7: Find the Time at Maximum Height

The maximum height is reached when the velocity is zero, i.e., \( v(t) = 0 \). Solve \( -32t + 300 = 0 \) for \( t \). This gives \( t = \frac{300}{32} = 9.375 \, \text{seconds}. \)

Step 8: Calculate Maximum Height

Substitute \( t = 9.375 \) into the height function \( s(t) = -16t^2 + 300t \): \( s(9.375) = -16(9.375)^2 + 300(9.375) = 1406.25 \, \text{feet}. \)

Key Concepts

Free-Fall Motion

Free-fall motion describes the motion of an object moving under the influence of gravity alone, without any other forces acting on it such as air resistance. In our exercise, the cannonball moving upwards after being shot is a perfect example of this concept, as the only force considered is gravity. This simulates an ideal condition where external factors, such as air friction, are ignored.

When in free-fall motion, the object initially moves upward against gravity until the force of gravity slows it down to a stop, then accelerates it downward. The motion can be analyzed using differential equations that connect time, velocity, and position of the object:

• The equation for acceleration is given by Newton's second law, which in this case is the force of gravity.
• The velocity of the object can be found by integrating the acceleration with respect to time.
• Similarly, the height at any point in time can be determined by integrating the velocity function over time.

Understanding free-fall motion is crucial as it forms the basis for solving many physics problems involving objects moving in a gravitational field. If you imagine the cannonball's entire trajectory, from launch to the highest point and back to the ground, that's the essence of free-fall motion in a gravitational field.

Initial Velocity

Initial velocity refers to the speed and direction at which an object starts its motion. In our case, the cannonball's initial velocity is given as 300 ft/s upwards. Initial velocity is critical because it sets the start conditions for the entire motion and influences the maximum height and duration of flight.

Here’s why initial velocity is important in free-fall calculations:

• The greater the initial velocity upward, the longer it will take before gravity brings the object to a stop and pulls it back down, increasing the maximum height reached.
• Initial velocity requires us to handle calculations in both vertical and horizontal components separately, but in this one-dimensional example, only the vertical component is relevant.
• In solving differential equations governing motion, initial velocity serves as the necessary initial condition to find particular solutions.

To calculate motion parameters like maximum height or time of flight, initial velocity becomes part of the equations derived from the laws of motion. For example, knowing initial velocity allowed us to find the constants in the velocity equation, leading to further solutions in the problem.

Gravitational Acceleration

Gravitational acceleration is the constant rate at which an object's velocity changes due to the force of gravity. On Earth, this is approximately 32 ft/s², as noted in the exercise. This value is paramount in calculating the trajectory and behavior of objects in free-fall.

Key facts about gravitational acceleration include:

• It always acts downward towards the Earth's center, which means it's a negative value when considering motion upwards.
• The value of gravitational acceleration is independent of the mass of the object; heavy and light objects fall at the same rate in the absence of air resistance.
• In the absence of other forces, the acceleration of gravity alone causes a linear decrease in velocity for objects traveling upwards until they stop, which then turns into an increase in velocity as they fall back down.

Using gravitational acceleration, you can predict the time it takes for an object to ascend and then fall back to the ground, or calculate its velocity and position at any given moment. It’s the fundamental piece in many physics equations and problems that involve motion caused by gravity, as shown in our cannonball example.