Question

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 small rocket is fired from a launch pad \(10 \mathrm{m}\) above the grout with an initial velocity, in \(\mathrm{m} / \mathrm{s}\), of (300,400,500) . A crosswi blowing to the north produces an acceleration of the rocket of \(25 \mathrm{m} / \mathrm{s}^{2}\)

Short answer

b. How would you describe the shape of the rocket's trajectory? c. What equations would you use to determine the time of flight and range of the rocket? d. How would you find the maximum height of the rocket?

Step-by-step solution

Find the velocity components in each direction

The initial velocity vector of the rocket is given as (300, 400, 500) m/s. We can separate this into the \(x\), \(y\) and \(z\) components, where \(v_x = 300\) m/s, \(v_y = 400\) m/s, and \(v_z = 500\) m/s.

Find the acceleration components in each direction

The rocket experiences a crosswind acceleration towards the north (\(y\)-direction) of 25 m/s². Therefore, we have acceleration components as follows: - \(x\)-direction: \(x\)-axis points east, and there is no acceleration acting in this direction, so \(a_x = 0\). - \(y\)-direction: \(y\)-axis points north, and the crosswind acceleration acts in this direction, so \(a_y = 25\) m/s². - \(z\)-direction: The only acceleration acting in the vertical direction is the gravitational force, which acts in the opposite direction of the positive \(z\)-direction, so \(a_z = -g\), where \(g = 9.81\) m/s².

Find the velocity and position vectors

To find the velocity vector at any time t, we can use the following equation for each component: $$ v_i(t) = v_{0i} + a_i t $$ where \(v_i\) is the velocity component in the \(i\)-direction (\(x, y, z\)), \(v_{0i}\) is the initial velocity component, and \(a_i\) is the acceleration in that direction. Thus, we have the following velocity components at any time t: - \(x\)-direction: \(v_x(t) = 300\) m/s. - \(y\)-direction: \(v_y(t) = 400 + 25t\) m/s. - \(z\)-direction: \(v_z(t) = 500 - 9.81t\) m/s. Now, to get the position vector at any time t, we can use the following equation for each component: $$ r_i(t) = r_{0i} + v_{0i} t + \frac{1}{2} a_i t^2 $$ where \(r_i\) is the position component in the \(i\)-direction and \(r_{0i}\) is the initial position component. In this problem, the rocket is fired from 10 meters above the ground, so \(r_{0z} = 10\) m and \(r_{0x} = r_{0y} = 0\). We thus have the following position components at any time t: - \(x\)-direction: \(r_x(t) = 300t\) m. - \(y\)-direction: \(r_y(t) = 400t + \frac{1}{2}(25)t^2\) m. - \(z\)-direction: \(r_z(t) = 10 + 500t - \frac{1}{2}(9.81)t^2\) m.

Make a sketch of the trajectory

You can either use plotting software or graph paper to sketch the trajectory of the rocket by plotting the position coordinates in 3D space at different time intervals. The path should resemble a parabola with a curvature in the \(y\)-direction due to the crosswind acceleration.

Determine the time of flight and range of the object

The time of flight can be determined by finding when the rocket reaches the ground (i.e., when \(r_z(t) = 0\)). Use the equation of motion in the \(z\)-direction to find this time \(t_f\). Once you have the time of flight, you can find the range of the object in the \(x\) and \(y\) directions, using the equations of motion as follows: - \(x\)-direction: \(R_x = r_x(t_f)\) - \(y\)-direction: \(R_y = r_y(t_f)\)

Determine the maximum height of the object

To determine the maximum height of the rocket, first find the time when the vertical velocity component becomes zero (\(v_z(t_h) = 0\)). This time, \(t_h\), will correspond to the highest point in the trajectory. After finding \(t_h\), substitute it into the \(z\)-direction position equation to obtain the maximum height of the rocket: $$ H = r_z(t_h) $$

Key concepts

Velocity Vector

In projectile motion, understanding the velocity vector is crucial to analyze the movement of an object. A velocity vector indicates both the speed and direction of motion. It is composed of three components, corresponding to each coordinate axis: \(x\), \(y\), and \(z\). In this scenario, a rocket has an initial velocity vector \((300, 400, 500)\) m/s. This means:

• \(v_x = 300\) m/s in the eastward direction.
• \(v_y = 400\) m/s in the northward direction, affected by a crosswind.
• \(v_z = 500\) m/s upward, countering gravity.

The velocity in each direction can change over time due to acceleration. The formulas used to find updated velocity include the initial velocity, acceleration, and time:\[v_i(t) = v_{0i} + a_i t\]This tells us how the speed of the rocket adjusts as it flies.

Position Vector

The position vector defines the location of an object at any given time in a 3D space, based on its motion parameters. Initially, our rocket starts at a height of 10 m above the ground and computes its position using:

• Initial position \((0, 0, 10)\) m.
• Account for the initial velocity along with time \(t\).
• Include any acceleration impacts over time.

The position vector components can be calculated as:\[r_i(t) = r_{0i} + v_{0i} t + \frac{1}{2} a_i t^2\]Thus:

• \(r_x(t) = 300t\)
• \(r_y(t) = 400t + \frac{1}{2}(25)t^2\)
• \(r_z(t) = 10 + 500t - \frac{1}{2}(9.81)t^2\)

This allows understanding of how the rocket's position changes over time in each direction, reflecting its trajectory.

Trajectory Sketch

A trajectory sketch provides a visual representation of an object's path in the space it moves through. For our rocket, it’s essential to plot the position vector at various time intervals to appreciate its journey. Due to the forces acting on it, the trajectory typically forms a parabolic shape:

• The parabola's vertex represents the highest point the rocket reaches.
• The crosswind slightly curves the path in the \(y\)-direction.
• Overall trajectory in a 3D space shows the changes over time through the \(x\), \(y\), and \(z\)-axes.

By sketching this, one can observe how the forces, like gravity and crosswind, influence the rocket’s motion, illustrating key points such as launch, peak height, and landing.

Time of Flight

The time of flight in projectile motion determines how long the object remains in air before touching the ground again. It is a vital aspect of projectile motion analysis. To find it, one needs to determine when the position in the \(z\)-direction becomes zero again \((r_z(t_f) = 0)\):

• Solve the equation for \(r_z(t)\) to find \(t_f\).
• Insert values into the equation to get the exact time it takes to land back.

This time decides how far the rocket has traveled horizontally (range) in the \(x\) and \(y\) axes. Understanding the time of flight helps visualize the projectile's trajectory and anticipate its landing point.

Maximum Height

Determining the maximum height is key to understanding the peak of the projectile's path, which is influenced by initial velocity and vertical acceleration. The point at which the vertical component of the velocity becomes zero represents the peak of the trajectory. This means:

• At maximum height, \(v_z(t_h) = 0\).
• Finding \(t_h\) enables computing the maximum height.

Once \(t_h\) is known, substitute it into the position formula for the \(z\)-direction:\[H = r_z(t_h)\]This calculation lets you find the maximum altitude the rocket achieves, offering a crucial parameter for analyzing projects and ensuring safe operations, especially in aerospace applications.