Problem 27 - Essential Calculus Early Transcendentals

Question

A ball is thrown eastward into the air from the origin (in the direction of the positive \(x\) -axis). The initial velocity is \(50 \mathbf{i}+80 \mathbf{k},\) with speed measured in feet per second. The spin of the ball results in a southward acceleration of \(4 \mathrm{ft} / \mathrm{s}^{2},\) so the acceleration vector is \(\mathbf{a}=-4 \mathrm{j}-32 \mathrm{k}\) Where does the ball land and with what speed?

Step-by-Step Solution

Answer

The ball lands at (250, -50, 0) and its speed is about 53.85 ft/s.

Step 1: Define Initial Quantities

We are given the initial velocity \(\mathbf{v}_0 = 50 \mathbf{i} + 80 \mathbf{k}\) and acceleration \(\mathbf{a} = -4 \mathbf{j} - 32 \mathbf{k}\). The initial position \(\mathbf{r}_0\) is the origin, so \(\mathbf{r}_0 = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}\).

Step 2: Determine the Velocity Function

The velocity function \(\mathbf{v}(t)\) can be found using the equation \(\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a}t\). Therefore, \(\mathbf{v}(t) = (50 \mathbf{i} + 80 \mathbf{k}) + (-4 \mathbf{j} - 32 \mathbf{k})t = 50 \mathbf{i} - 4t \mathbf{j} + (80 - 32t) \mathbf{k}\).

Step 3: Calculate the Position Function

The position function \(\mathbf{r}(t)\) is given by the integral of velocity \(\mathbf{v}(t)\). Integrating each component, we have \(\mathbf{r}(t) = \int \{50 \mathbf{i} - 4t \mathbf{j} + (80 - 32t) \mathbf{k}\} dt = (50t) \mathbf{i} - 2t^2 \mathbf{j} + (80t - 16t^2) \mathbf{k} + \mathbf{C}\). The constant \(\mathbf{C}\) is \(0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}\) because the initial position is the origin.

Step 4: Set K-component to Zero to Find Time of Landing

To find when the ball lands, set the \(\mathbf{k}\) component to zero: \(80t - 16t^2 = 0\). Solving for \(t\), we factor out \(16t\) to get \(16t(5 - t) = 0\), so \(t = 0\) or \(t = 5\). The ball lands after 5 seconds.

Step 5: Find the Landing Position

Substitute \(t = 5\) into \(\mathbf{r}(t)\) to find the position: \(\mathbf{r}(5) = (50 \times 5) \mathbf{i} - 2(5)^2 \mathbf{j} + (80 \times 5 - 16 \times 5^2) \mathbf{k}\). Simplifying, \(\mathbf{r}(5) = 250 \mathbf{i} - 50 \mathbf{j} + 0 \mathbf{k}\). Thus, the ball lands at the position \( (250 \mathbf{i}, -50 \mathbf{j}, 0 \mathbf{k}) \).

Step 6: Calculate the Speed upon Landing

Find speed by substituting \(t = 5\) into \(\mathbf{v}(t)\): \(\mathbf{v}(5) = 50 \mathbf{i} - 4(5) \mathbf{j} + (80 - 32(5)) \mathbf{k} = 50 \mathbf{i} - 20 \mathbf{j}\). The speed is \(\|\mathbf{v}(5)\| = \sqrt{50^2 + (-20)^2} = \sqrt{2500 + 400} = \sqrt{2900}\). Thus, the speed is approximately 53.85 feet per second.

Key Concepts

Vectors

Vectors are mathematical objects used to represent quantities that have both direction and magnitude. In projectile motion, like the scenario with the ball being thrown, vectors efficiently describe quantities such as position, velocity, and acceleration.

• Position Vector: Indicates the ball's location in space at any given time, starting from a defined origin. In this problem, it is initially zero since the ball starts at the origin.
• Velocity Vector: Specifies how rapidly and in what direction the ball's position changes with time. Initially given as \(50\mathbf{i} + 80\mathbf{k}\), it blends directional speed along the east (\(\mathbf{i}\)) and upwards (\(\mathbf{k}\)) axes.
• Acceleration Vector: Describes how the velocity of the ball changes, incorporating effects like gravity and any added forces, such as the southward spin-induced acceleration in this case.

Understanding these vectors' directions and magnitudes is crucial for accurately mapping the trajectory and final position of the ball.

Velocity Function

The velocity function is a mathematical representation of how an object's velocity changes over time, crucial for predicting the object's future motion in projectile problems.

• Initial Velocity: Here, the ball's initial velocity is a vector described by \(50\mathbf{i} + 80\mathbf{k}\), indicating its initial eastward and upward movement.
• Velocity Function Formula: To find the velocity at any time \(t\), use \( \mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a}t\). This adjusts initial velocity \(\mathbf{v}_0\) by the effects of acceleration over time.
• Calculation: In this exercise, substituting the acceleration \(-4\mathbf{j} - 32\mathbf{k}\) provides \( \mathbf{v}(t) = 50 \mathbf{i} - 4t \mathbf{j} + (80 - 32t) \mathbf{k}\).

The velocity function is pivotal for determining speed and direction, particularly when assessing the ball's impact speed.

Position Function

The position function is derived by integrating the velocity function over time, showing where an object is located within a plane or space at any given moment.

• Integration Process: Integrate each component of the velocity vector to set up the position function. For this problem, \( \mathbf{r}(t) = \int \{50 \mathbf{i} - 4t \mathbf{j} + (80 - 32t) \mathbf{k}\} dt\).
• Constants of Integration: In projectile problems starting from the origin, the integration constant \(\mathbf{C}\) is typically zero, reflecting the initial starting point.
• Final Position Expression: This evaluates to \[ \mathbf{r}(t) = (50t) \mathbf{i} - 2t^2 \mathbf{j} + (80t - 16t^2) \mathbf{k} \].

Given \(t=5\), this equation gives the precise landing point of the ball when substituting into its resolved form.

Integration

Integration is a fundamental concept in calculus used to determine the position function from its derivative, the velocity function. By integrating, you essentially accumulate values over an interval, like adding numerous tiny velocity vectors to depict position.

• Basic Mechanics: In this context, integration is hauling together the velocity components to track overall displacement.
• Component-wise Integration: Each vector component is treated individually, which can sometimes involve quadratic and other simple polynomial calculations, as with \(-4t \mathbf{j}\) integrating to \(-2t^2 \mathbf{j}\).
• Initial Conditions: The integration constant is usually addressed by initial conditions, here easy since the ball starts precisely at the origin, confirming zero added displacement at \(t=0\).

Mastering integration can feel challenging at first, but remembering it's essentially accumulating changes over time can simplify this vital process in physics.