Problem 11 - Advanced Engineering Mathematics

Question

A shell is fired from ground level with an initial speed of \(480 \mathrm{ft} / \mathrm{s}\) at an angle of elevation of \(30^{\circ}\). Find: (a) a vector function and parametric equations of the shell's trajectory, (b) the maximum altitude attained, (c) the range of the shell, and (d) the speed at impact.

Step-by-Step Solution

Answer

The maximum altitude is 900 ft, the range is 6235 ft, and the impact speed is 480 ft/s.

Step 1: Understand the Problem

We are given initial conditions: initial speed \(v_0 = 480 \, \text{ft/s}\), and the angle of launch \( \theta = 30^{\circ} \). We need to find a vector function for the trajectory, parametric equations, maximum altitude, range, and speed at impact. Use gravitational acceleration \( g = 32 \, \text{ft/s}^2 \).

Step 2: Write Vector Function of the Trajectory

The initial velocity vector is \( \mathbf{v_0} = 480 \langle \cos(30^{\circ}), \sin(30^{\circ}) \rangle \). Compute the components: \(v_{0x} = 480 \times \cos(30^{\circ}) = 240\sqrt{3} \, \text{ft/s}\) and \(v_{0y} = 480 \times \sin(30^{\circ}) = 240 \, \text{ft/s}\). The vector function is:\[ \mathbf{r}(t) = (240\sqrt{3}t)\mathbf{i} + (240t - 16t^2)\mathbf{j} \]

Step 3: Find Parametric Equations

Parametrize the vector function. The parametric equations are:\[ x(t) = 240\sqrt{3}t \]\[ y(t) = 240t - 16t^2 \]

Step 4: Calculate Maximum Altitude

The maximum height occurs when the vertical velocity is zero: \(v_{y}(t) = 240 - 32t = 0\). Solving gives \(t = 7.5\) seconds. Plug \(t = 7.5\) into \(y(t)\):\[ y(7.5) = 240 \times 7.5 - 16 \times (7.5)^2 = 900 \text{ ft} \]

Step 5: Determine the Range of the Shell

At impact, the shell returns to ground level: \(y(t) = 0\). Solve \(240t - 16t^2 = 0\), which gives roots \(t = 0\) and \(t = 15\) seconds. The range is calculated by plugging \(t = 15\) into \(x(t)\):\[ x(15) = 240\sqrt{3} \times 15 = 3600\sqrt{3} \approx 6235 \text{ ft} \]

Step 6: Calculate Speed at Impact

Find the velocity vector at \(t = 15\):\[ v_x = 240\sqrt{3}, \quad v_y = 240 - 32 \times 15 = -240 \, \text{ft/s} \]The speed is:\[ v = \sqrt{(240\sqrt{3})^2 + (-240)^2} = 480 \, \text{ft/s} \]

Step 7: Wrap Up Results

We've derived the trajectory vector function, parametric equations, maximum altitude of 900 ft, a range of approximately 6235 ft, and an impact speed of 480 ft/s.

Key Concepts

Vector Functions

When studying projectile motion, vector functions provide a method to describe the path of an object, like a shell fired from a cannon, in terms of vectors. This is particularly useful because vectors allow us to break down the motion into horizontal and vertical components. The initial velocity vector, allowing us to specify both direction and magnitude, becomes essential in these calculations. In our problem, the initial velocity vector is composed of two components: horizontal and vertical. The horizontal component is calculated using the cosine of the launch angle, while the vertical component is derived using the sine of the angle. This results in a vector function that mathematically represents the path of the shell:

• Horizontal component: \( v_{0x} = 240\sqrt{3} \, \text{ft/s} \)
• Vertical component: \( v_{0y} = 240 \, \text{ft/s} \)

These components are then used to define the vector function \( \mathbf{r}(t) = (240\sqrt{3}t)\mathbf{i} + (240t - 16t^2)\mathbf{j} \), which describes the trajectory of the projectile as a function of time \( t \).

Parametric Equations

Parametric equations are used to express the coordinates of a projectile as functions of time. This approach allows us to easily extract information about the position at any given time along its path. They typically separate the motion into its constituent parts: one equation for the horizontal (x-axis) component and another for the vertical (y-axis) component. For our fired shell, these equations are derived from the earlier vector function.

• Horizontal position equation: \( x(t) = 240\sqrt{3}t \)
• Vertical position equation: \( y(t) = 240t - 16t^2 \)

These equations enable you to find out the exact position of the shell at any moment during its flight. Note that the vertical equation incorporates the effect of gravity, represented as \( -16t^2 \), which gradually decreases the vertical position over time until the projectile touches the ground.

Maximum Altitude

The maximum altitude, or the highest point reached by a projectile during its motion, occurs when the vertical component of its velocity becomes zero. This is because at the highest point, the projectile stops rising and starts descending. To find this moment, we set the derivative of the vertical position function, which is the vertical velocity, to zero and solve for the time. In this case:\[ v_y(t) = 240 - 32t = 0 \]From this equation, we find that the time \( t \) is 7.5 seconds. Substituting back into the vertical position equation \( y(t) \), we calculate the maximum altitude:\[ y(7.5) = 240 \times 7.5 - 16 \times (7.5)^2 = 900 \, \text{ft} \]This result tells us that the shell reaches a peak altitude of 900 feet before coming down.

Range of Projectile

The range of a projectile refers to the horizontal distance it covers from the point of launch to the point of impact, assuming it lands at the same vertical level from which it started. For our shell, the range is determined by finding the time it takes to return to ground level, meaning the vertical position \( y(t) \) is zero. Solving this for time gives us the total time of flight.We solve the equation:\[ 240t - 16t^2 = 0 \]This provides the flight duration with roots at \( t = 0 \) and \( t = 15 \) seconds, indicating the projectile is in the air for 15 seconds. We then use this time in the horizontal position equation to find the range:\[ x(15) = 240\sqrt{3} \times 15 = 3600\sqrt{3} \approx 6235 \, \text{ft} \]Thus, the shell travels approximately 6235 feet horizontally before coming to rest at the same level as it was fired.