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 bullet is fired from a rifle 1 m above the ground in a northeast. direction. The initial velocity of the bullet is \((200,200,0) \mathrm{m} / \mathrm{s}\)

Short answer

In summary, for a bullet fired from a rifle 1 meter above the ground with an initial velocity vector of (200, 200, 0) m/s, the position vector is given by \(\vec{r}(t) = 200t\hat{i} + 200t\hat{j} + (1 - 4.9t^2)\hat{k}\), and the velocity vector is given by \(\vec{v}(t) = 200\hat{i} + 200\hat{j} - 9.8t\hat{k} \, \mathrm{m/s}\). The bullet follows a parabolic trajectory in 3D space. The time of flight is approximately 0.452 seconds, the range is approximately 127.9 meters, and the maximum height reached is 1 meter.

Step-by-step solution

Calculate the horizontal components of the velocity and position vectors

Since the initial velocity vector has components in both the \(x\) and \(y\) directions, we can write it as \(\vec{v_0} = (200,200,0) \mathrm{m/s}\). The position vector \(\vec{r}\) for any time \(t\) in 3 dimensions is given as: \(\vec{r}(t) = \vec{r_0} + \vec{v_0}t + \frac{1}{2}\vec{g}t^2\) Given the ground is horizontal, we have \(x\) and \(y\) components as the horizontal components of motion, so our initial position vector is: \(\vec{r_0} = (0,0,1) \mathrm{m}\) and \(\vec{g} = (0,0,-9.8 \, \mathrm{m}/\mathrm{s^2})\) for gravity. Our horizontal components of the position vector are: \(\vec{r}_{x}(t) = \vec{r}_{0_{x}} + v_{0_x}t\) \(\vec{r}_{x}(t) = 200t\hat{i}\) \(\vec{r}_{y}(t) = \vec{r}_{0_{y}} + v_{0_y}t\) \(\vec{r}_{y}(t) = 200t\hat{j}\)

Calculate the vertical component of the position vector

Now we find the vertical component of the position vector: \(\vec{r}_{z}(t) = (\vec{r}_{0_{z}} + v_{0_z}t + \frac{1}{2}g_zt^2)\) \(\vec{r}_{z}(t) = (1 + 0 + \frac{1}{2}(-9.8)t^2)\hat{k}\) \(\vec{r}_{z}(t) = (1 - 4.9t^2)\hat{k}\)

Combine position vector components

Now, we can combine the \(x\), \(y\), and \(z\) components of the position vector: \(\vec{r}(t) = 200t\hat{i} + 200t\hat{j} + (1 - 4.9t^2)\hat{k}\)

Calculate velocity vector components

Next, we find the velocity vector components in each direction. The horizontal components undergo no acceleration due to gravity, so their velocities remain constant: \(\vec{v}_{x}(t) = v_{0_x}\) \(\vec{v}_{x}(t) = 200 \mathrm{m/s}\) \(\vec{v}_{y}(t) = v_{0_y}\) \(\vec{v}_{y}(t) = 200 \mathrm{m/s}\) For the vertical component, we find: \(\vec{v}_{z}(t) = v_{0_z} + g_zt\) \(\vec{v}_{z}(t) = -9.8t\hat{k}\)

Combine velocity vector components

Combining the velocity components, we find: \(\vec{v}(t) = 200\hat{i} + 200\hat{j} - 9.8t\hat{k} \, \mathrm{m/s}\)

Sketch the trajectory

Since the horizontal components of the motion are constant, the trajectory of the bullet in the \(x\)-\(y\) plane will be a straight line. The vertical component undergoes a parabolic behavior as it is affected by gravity. Therefore, the trajectory of the bullet will be a parabola in 3D space.

Calculate time of flight and range

The time of flight can be calculated from the vertical component of the position vector, when \(\vec{r}_z(t) = 0\): \(0 = 1 - 4.9t^2\) \(t^2 = \frac{1}{4.9}\) \(t = \sqrt{\frac{1}{4.9}} \approx 0.452 \, \mathrm{s}\) During this time, the horizontal components \(\vec{r}_x\), and \(\vec{r}_y\) traveled: \(R_x = 200t = 200 \times 0.452 \approx 90.4 \, \mathrm{m}\) \(R_y = 200t = 200 \times 0.452 \approx 90.4 \, \mathrm{m}\) So, the range of the bullet is \(\sqrt{R_x^2 + R_y^2} \approx 127.9 \, \mathrm{m}\).

Calculate the maximum height

To find the maximum height reached by the bullet, we need to find the vertex of the parabola in its vertical motion: \(t_{vertex} = \frac{-v_{0_z}}{2g_z} = 0\) \(h_{max} = 1 - 4.9(0)^2 = 1 \, \mathrm{m}\) The maximum height reached by the bullet is 1 meter.

Key concepts

Position Vectors

In projectile motion, position vectors help us understand the location of an object in space at any given time. When an object is fired or thrown, like a bullet, it moves along a certain path. We express this path using coordinates relative to a point in space. These coordinates give us the position vector, which in three dimensions is written as \(\vec{r}(t)\).
For our bullet example, the initial position was set at 1 meter above the ground and its position vector can be described as follows:

• Horizontal components: \(\vec{r}_{x}(t) = 200t\hat{i}\) and \(\vec{r}_{y}(t) = 200t\hat{j}\), reflecting the bullet’s movement in the east and north directions.
• Vertical component: \(\vec{r}_{z}(t) = 1 - 4.9t^2\) showing the up-and-down motion due to gravity.

These position vectors collectively help us track where the bullet is at any given time \(t\).

Velocity Vectors

Velocity vectors indicate how fast and in which direction an object moves. They are crucial in understanding the dynamics of projectile motion. Originally, the bullet was fired with an initial velocity vector, \(\vec{v_0} = (200, 200, 0)\; \mathrm{m/s}\), indicating simultaneous movement east (\(x\)) and north (\(y\)) but no initial vertical speed (\(z\)).
For the bullet's motion, the velocity vector changes with time like this:

• Horizontal velocities remain unchanged: \(\vec{v}_{x}(t) = 200\; \mathrm{m/s}\) and \(\vec{v}_{y}(t) = 200\; \mathrm{m/s}\).
• Vertical velocity is affected by gravity: \(\vec{v}_{z}(t) = -9.8t\hat{k}\), showing that gravity slows the bullet's ascent and accelerates its descent.

The velocity vector \(\vec{v}(t) = 200\hat{i} + 200\hat{j} - 9.8t\hat{k}\) helps predict how the bullet's speed and direction change over time.

Gravitational Force

Gravitational force is one of the main influences on an object's motion, particularly in the vertical direction. It is the downward force acting on the bullet, affected only by gravity, which causes it to follow a parabolic trajectory. In physics, this force is characterized by the acceleration due to gravity, represented as \(\vec{g} = (0, 0, -9.8)\; \mathrm{m/s^2}\).
Key aspects of gravitational force influencing the projectile motion include:

• Causing the vertical component of the bullet's position to change: As captured by \(\vec{r}_{z}(t) = 1 - 4.9t^2\), indicating the bullet falls downward over time.
• Affecting vertical velocity: The downward acceleration results in \(\vec{v}_{z}(t) = -9.8t\hat{k}\), pulling the bullet towards the ground.

Gravitational force plays a silent but continuous role, crucial in predicting how high the bullet will go and when it will hit the ground.

Trajectories

The trajectory is the path that an object follows through space as a function of time. When discussing projectile motion, the trajectory is often a parabolic curve. In the context of our bullet, its trajectory combines both its horizontal and vertical movements.
The shape and direction of the trajectory are dictated by:

• Initial velocity: Determines how far and fast the bullet moves horizontally and vertically.
• Gravity: Affects the vertical path making the trajectory parabolic.

The calculated components led us to deduce that the bullet travels in a northeast direction in a straight line within the horizontal plane (\(x\)-\(y\)) and describes a parabola vertically with decreasing height until it falls back to the ground. This understanding of trajectories is essential for predicting the overall path and final landing point of our projectile, which for the bullet, concludes approximately 127.9 meters from the launch point.