Problem 34 - Advanced Engineering Mathematics

Question

(a) Integrating \(d^{2} s / d t^{2}=-g\) we get \(v(t)=d s / d t=-g t+c .\) From \(v(0)=300\) we find \(c=300,\) and we are given \(g=32,\) so the velocity is \(v(t)=-32 t+300\). (b) Integrating again and using \(s(0)=0\) we get \(s(t)=-16 t^{2}+300 t .\) The maximum height is attained when \(v=0,\) that is, at \(t_{a}=9.375 .\) The maximum height will be \(s(9.375)=1406.25 \mathrm{ft}\).

Step-by-Step Solution

Answer

Maximum height is 1406.25 ft at 9.375 seconds.

Step 1: Understand the Differential Equation

We start with \[ \frac{d^2 s}{dt^2} = -g \]This equation signifies that the second derivative of displacement \(s\) with respect to time \(t\) is equal to the negative of acceleration due to gravity \(g\).

Step 2: Integrate the Acceleration to Find Velocity

Integrate the equation \[ \frac{d^2 s}{dt^2} = -g \]with respect to \(t\) to find the velocity \(v(t)\):\[ v(t) = \int (-g) \, dt = -gt + c \]where \(c\) is the constant of integration.

Step 3: Determine the Constant for Velocity

We use initial condition \(v(0) = 300\) to find \(c\):\[ 300 = -g(0) + c \]Therefore, \(c = 300\).

Step 4: Substitute Values into Velocity Equation

With \(g = 32\) and \(c = 300\), we have:\[ v(t) = -32t + 300 \]

Step 5: Integrate Velocity to Find Displacement

Integrate the velocity equation to find displacement \(s(t)\):\[ s(t) = \int (-32t + 300) \, dt = -16t^2 + 300t + k \]where \(k\) is another constant of integration.

Step 6: Determine Constant for Displacement

Using initial condition \(s(0) = 0\), we find \(k\):\[ 0 = -16(0)^2 + 300(0) + k \]Thus, \(k = 0\).

Step 7: Determine Time for Maximum Height

The maximum height is reached when \(v(t) = 0\):\[ -32t + 300 = 0 \]\[ t = 9.375 \]

Step 8: Calculate Maximum Height

Substitute \(t = 9.375\) into \(s(t) = -16t^2 + 300t\):\[ s(9.375) = -16(9.375)^2 + 300(9.375) = 1406.25 \text{ ft} \]

Key Concepts

Kinematics in Physics

Kinematics is a branch of physics that describes the motion of objects. It focuses on quantities such as displacement, velocity, and acceleration without considering their causes. In this exercise, kinematics plays a crucial role as we explore the motion of a projectile under the influence of gravity.

When an object is in free fall, it experiences constant acceleration due to gravity, denoted by the symbol \( g \). This occurs because gravity is pulling the object downwards. We typically take \( g \) to be 32 ft/s² when dealing with problems in English units.

The exercise begins by using the differential equation \( \frac{d^2 s}{dt^2} = -g \), which states that the second derivative of position \( s \) with respect to time \( t \) is equal to the negative of the gravitational acceleration, indicating a downward acceleration.

Integration Techniques

Integration is a fundamental mathematical tool used to find functions from their derivatives. It is particularly useful in problems involving motion, where one can determine velocity from acceleration or displacement from velocity. In this exercise, we first integrate the acceleration \( d^2 s/d t^2 = -g \) to find velocity.

Starting with the integral of acceleration, we obtain the velocity equation: \( v(t) = \int (-g) \, dt = -gt + c \). Here, \( c \) is the constant of integration, which can be determined by initial conditions. In our case, the initial condition \( v(0) = 300 \) helps find that \( c = 300 \).

Next, integration of the velocity equation provides us the displacement, \( s(t) = \int (-32t + 300) \, dt = -16t^2 + 300t + k \). Applying the initial condition \( s(0) = 0 \) guides us in finding that \( k = 0 \).

• Step 1: Integrate acceleration to find velocity.
• Step 2: Use known values (initial conditions) to find constants \( c \) or \( k \).
• Step 3: Integrate velocity to find displacement.

Initial Conditions

Initial conditions are crucial in solving differential equations as they allow us to determine the constants of integration. In physics, these conditions are often initial values of physical quantities like position or velocity when time \( t = 0 \).

In our problem, two initial conditions are given:

• The initial velocity \( v(0) = 300 \) ft/s, which helps find the constant \( c \) in the velocity equation.
• The initial displacement \( s(0) = 0 \) helps find the constant \( k \) in the displacement equation.

These conditions reflect the state of the projectile at the beginning of its motion. They effectively "anchor" the mathematical equations to the physical scenario, allowing for meaningful and accurate calculations of velocity and displacement.

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. In this exercise, the projectile in discussion is initially moving with a velocity of 300 ft/s upwards.

The equations derived through differential equations and integration describe the trajectory of the projectile. The velocity equation \( v(t) = -32t + 300 \) and displacement \( s(t) = -16t^2 + 300t \) show how these quantities change over time.

A specific point of interest in projectile motion is the maximum height. It is reached when the velocity becomes zero, meaning that the projectile momentarily stops moving upwards before gravity pulls it back down. We set \( v(t) = 0 \) and solve for \( t \) to find that the maximum height occurs at \( t = 9.375 \) seconds. Placing this value back into the displacement equation, we calculate the maximum height to be 1406.25 feet.

• Initial upward velocity: 300 ft/s.
• Maximum height occurs when \( v(t) = 0 \).
• Calculated height: 1406.25 feet.

Each part of these calculations helps us fully understand the dynamics of projectile motion.