Question

Three-dimensional motion Consider the motion of the following objects. Assume the \(x\) -axis points east, the \(y\) -axis points north, the positive z-axis is vertical and opposite g. the ground is horizontal, and only the gravitational force acts on the object unless otherwise stated. a. Find the velocity and position vectors, for \(t \geq 0\). b. Make a sketch of the trajectory. c. Determine the time of flight and range of the object. d. Determine the maximum height of the object. A golf ball is hit east down a fairway with an initial velocity of \(\langle 50,0,30\rangle \mathrm{m} / \mathrm{s} .\) A crosswind blowing to the south produces an acceleration of the ball of \(-0.8 \mathrm{m} / \mathrm{s}^{2}\).

Short answer

Answer: At the time of maximum height (approx. 3.06 s), the position vector of the golf ball is approximately ⟨ 153, -3.76, 45.88 ⟩ m.

Step-by-step solution

Find the velocity and position vectors for all t ≥ 0.

First, analyze the accelerations in each direction. Gravitational acceleration is straight down, so it will affect the z-component with \(-g = -9.8 \mathrm{m} / \mathrm{s}^{2}\). The crosswind affects the south direction, which is the y-component, producing an acceleration of \(-0.8 \mathrm{m} / \mathrm{s}^{2}\). There's no acceleration in the x direction. So, the total acceleration vector is: $$ \vec{a}(t) = \langle0, -0.8, -9.8\rangle \mathrm{m} / \mathrm{s}^{2} $$ To find the velocity vectors as a function of time, integrate the acceleration components with respect to time: $$ \vec{v}(t) = \langle \int{0 dt},\int{-0.8 dt}, \int{-9.8 dt}\rangle + \vec{v_0} $$ Here, \(\vec{v_0}\) is the initial velocity vector \(\langle 50,0,30 \rangle \mathrm{m}\mathrm{s}^{-1}\). Therefore, the velocity vector is: $$ \vec{v}(t) = \langle 50,\ -0.8t,\ 30-9.8t\rangle \mathrm{m} / \mathrm{s} $$ To find the position vectors as a function of time, integrate the velocity components with respect to time: $$ \vec{r}(t) = \langle \int{50dt},\int{-0.8t dt}, \int{(30-9.8t) dt}\rangle + \vec{r_0} $$ Considering \(\vec{r_0} = \langle 0,0,0 \rangle \mathrm{m}\) (the golf ball is hit from the ground, and from the point of origin), we get: $$ \vec{r}(t) = \langle 50t,\ -0.4t^2,\ 30t-4.9t^2\rangle \mathrm{m} $$

Make a sketch of the trajectory.

To sketch the trajectory, plot the points in a 3D space using the position vectors calculated in step 1 as the coordinates. You would see a parabolic trajectory with a slight downward curve towards the south due to the effect of the crosswind.

Determine the time of flight and range of the object.

The time of flight is the time from when the ball is hit until it reaches the ground, which happens when the z-component of the position vector is equal to 0: $$ 30t - 4.9t^2 = 0 $$ Factoring out t, we get: $$ t(30 - 4.9t) = 0 $$ There are two solutions: t=0 (initial position) and \(t = \frac{30}{4.9} \approx 6.12 \mathrm{s}\) (time of flight) Now, find the range by calculating the position vector components at the time of flight: $$ \vec{r}(6.12) = \langle 50 \times 6.12,\ -0.4\times(6.12)^2,\ 0\rangle \mathrm{m} = \langle 306, -15.07, 0\rangle \mathrm{m} $$ The range of the object in the x, y and z directions would be 306 m, -15.07 m, and 0 m respectively.

Determine the maximum height of the object.

To find the maximum height, find the time when the z-component of the velocity vector is equal to 0: $$ 30 - 9.8t = 0 $$ Solve for t: $$ t \approx 3.06 \mathrm{s} $$ Now, calculate position vector components at this time: $$ \vec{r}(3.06) = \langle 50 \times 3.06,\ -0.4\times(3.06)^2,\ 30\times3.06 - 4.9\times(3.06)^2\rangle \mathrm{m} $$ The maximum height in the z-direction is approximately 45.88 m.

Key concepts

Velocity Vector

In three-dimensional motion, the velocity vector represents the rate of change of the position vector with respect to time.
It includes components for each axis of motion—namely, the x, y, and z axes.
For the golf ball's motion described in the problem, the velocity vector is given by:

• The x-component, which remains constant at 50 m/s due to no acceleration in that direction.
• The y-component, which changes over time due to the south-ward crosswind effect, modeled as offset by the acceleration from the crosswind: y-component: \(-0.8t \, \text{m/s} \)
• The z-component, affected by gravity, leading to \(30 - 9.8t \, \text{m/s} \)representing vertical motion.

Knowing how these components interact helps in predicting the ball's speed and direction at any given time.

Position Vector

The position vector in three-dimensional space defines where the particle is located at any instant.
It also evolves based on the velocity vector by the integration over time.
The position vector for the golf ball provides its location in terms of the following:

• X-component, \(50t \, \text{meters} \), indicating constant movement eastward at 50 m/s.
• Y-component, \(-0.4t^2 \, \text{meters} \), affected by the crosswind, showing a gradual southward movement.
• Z-component, \(30t - 4.9t^2 \, \text{meters} \), representing vertical displacement controlled by initial upward velocity and gravitational pull.

This describes the golf ball's continuous journey through space, facilitating trajectory plotting.

Trajectory

The trajectory of an object in three-dimensional motion describes its path through space over time.
Understanding the trajectory gives insights into both the shape of the path and the influences acting on the object.
For the golf ball:

• The x component ensures it moves steadily eastward.
• The y component's interaction with the southern crosswind results in a curling path to the south.
• The z component gives the classic projectile parabolic arc, reaching an apex before descending.

Albeit impacted by various forces, plotting the trajectory confirms how motion progresses and predicts future movement. Piecing together the trajectory from the position vectors shows a smooth sweeping curve characterized by eastward progress and moderate southward drift.

Time of Flight

The time of flight is how long an object remains in motion, often gauged by how long it takes to return to ground level.
For the golf ball, time of flight ends when its z-component of the position vector returns to zero.
This happens when the ball has completed its ascent and descent cycle.Setting \( 30t - 4.9t^2 = 0 \) and solving provides the times for the beginning and end:

• Initial time, \(t = 0\) sec, is when the ball is hit.
• Final time, \(t \approx 6.12\) sec, marks when the ball hits the ground again.

Understanding time of flight helps determine how far and high the ball travels during its flight.

Maximum Height

Maximum height reflects the highest vertical point reached in a projectile's motion.
For the golf ball, it happens when vertical velocity momentarily reaches zero as the ball reverses direction to fall back down.
To find it, solve:\( 30 - 9.8t = 0 \) for \( t \). It reveals the time at which maximum height occurs:

• \( t \approx 3.06 \, \text{seconds} \)

Substitute this time back into the position vector to find maximum height in the z-direction:

• At \( t \approx 3.06 \, \text{seconds}, z \approx 45.88 \, \text{meters} \).

Recognizing it gives one a full understanding of how gravity and initial velocity interplay to limit upward motion.