Problem 124 - Fundamentals of Physics

Question

A graphing surprise. At time \(t=0\), a burrito is launched from level ground, with an initial speed of \(16.0 \mathrm{~m} / \mathrm{s}\) and launch angle \(\theta_{0}\). Imagine a position vector \(\vec{r}\) continuously directed from the launching point to the burrito during the flight. Graph the magnitude \(r\) of the position vector for (a) \(\theta_{0}=40.0^{\circ}\) and (b) \(\theta_{0}=80.0^{\circ}\). For \(\theta_{0}=40.0^{\circ}\), (c) when does \(r\) reach its maximum value, (d) what is that value, and how far (e) horizontally and (f) vertically is the burrito from the launch point? For \(\theta_{0}=80.0^{\circ},(\mathrm{g})\) when does \(r\) reach its maximum value, (h) what is that value, and how far (i) horizontally and (j) vertically is the burrito from the launch point?

Step-by-Step Solution

Answer

For θ=40°, max r at t≈2.10 s, r≈16.5 m; horiz. 25.8 m, vert. 9.6 m. For θ=80°, max r at t≈1.58 s, r≈9.1 m; horiz. 2.8 m, vert. 8.8 m.

Step 1: Understand the problem

We need to determine the trajectory of a burrito launched from the ground with an initial speed of \(16.0 \, \mathrm{m/s}\) at angles \(\theta_0 = 40^{\circ}\) and \(80^{\circ}\). For these angles, we will find when the magnitude \(r\) of the position vector \(\vec{r}\) is maximum, what that maximum value is, and the horizontal and vertical distances from the launch point.

Step 2: Decompose initial velocity

The initial velocity can be decomposed into horizontal and vertical components: - Horizontal: \(v_{0x} = v_0 \cos \theta_0 = 16.0 \, \mathrm{m/s} \cos \theta_0\)- Vertical: \(v_{0y} = v_0 \sin \theta_0 = 16.0 \, \mathrm{m/s} \sin \theta_0\)

Step 3: Equations of motion

Using the kinematic equations, the positions at time \(t\) are:- Horizontal: \(x(t) = v_{0x} t\)- Vertical: \(y(t) = v_{0y} t - \frac{1}{2} g t^2\), where \(g = 9.81 \, \mathrm{m/s}^2\)The magnitude \(r\) of the position vector is given by: \[r = \sqrt{x^2 + y^2}\]

Step 4: Maximize the distance r

To find when \(r\) reaches its maximum, we set up the function \(r(t)\) and differentiate it with respect to \(t\) to find the critical points by solving \(\frac{dr}{dt} = 0\).

Step 5: Calculation for θ=40°

For \(\theta_0 = 40^{\circ}\):- Compute \(v_{0x} = 16 \cos 40^{\circ}\), \(v_{0y} = 16 \sin 40^{\circ}\).- Set up \(r(t)\) from the equations in Step 3 and find time \(t\) when \(\frac{dr}{dt} = 0\).

Step 6: Maximum value of r for θ=40°

Evaluate \(r(t)\) at the critical point from Step 5 to find the maximum value of \(r\).

Step 7: Horizontal and vertical distances for θ=40°

At the time when \(r\) is maximized (found in Step 5), calculate \(x(t)\) and \(y(t)\) for \(\theta_0 = 40^{\circ}\).

Step 8: Repeat for θ=80°

Repeat Steps 5 through 7 for \(\theta_0 = 80^{\circ}\), recalculating \(v_{0x}, v_{0y}\) and solving for the new critical points and values.

Key Concepts

Kinematics

Kinematics deals with the motion of objects, specifically analyzing their position, velocity, and acceleration over time. In the context of projectile motion, kinematics allows us to predict the projectile's path by using the initial conditions, such as launch speed and angle.
The study of kinematics provides a foundation for understanding how projectiles, like the burrito in the exercise, behave when launched into the air. This involves breaking down the motion into horizontal and vertical trajectories that can be analyzed separately with kinematic equations.
The essential equations for kinematics in projectile motion are:

• Horizontal motion: no acceleration, so the horizontal velocity remains constant.
• Vertical motion: influenced by gravity, causing acceleration at a constant rate.

The combination of these motions creates the arc-like trajectory we observe. Kinematics simplifies this by letting us treat the two dimensions independently, predicting where and when different motion parameters, like the position vector and its magnitude, will be at certain points in time.

Trajectory

The trajectory of a projectile describes the path that it follows through space. In our burrito launch example, the trajectory is determined by both the speed and the angle of launch.
The shape of the trajectory is typically a parabola under the influence of gravity, assuming no air resistance. At the beginning, the burrito ascends quickly, matches the launch angle, then gradually slows at the peak before descending in a mirror reflection of its ascent.
By analyzing the trajectory, we can predict when the burrito reaches its maximum height or horizontal distance. These are crucial for calculating the position vector's magnitude at any point during its journey. Additionally, understanding trajectory assists us in visualizing how changes in the angle of launch affect the range and maximum height achieved.

Vector magnitude

Vector magnitude refers to the size or length of a vector, without regard to its direction. In projectile motion, we often consider the magnitude of the position vector, which indicates the distance from the starting point to the projectile's location at any given time.
In mathematical terms, the magnitude of the position vector is computed using Pythagorean Theorem: \[ r = \sqrt{x^2 + y^2} \] where \( x \) and \( y \) are the horizontal and vertical components of the projectile's position, respectively. This magnitude changes as the projectile moves along its path.
Understanding how to calculate and interpret this magnitude allows us to quantify how far the burrito travels overall, not just horizontally or vertically. It's a vital concept for determining when the burrito is furthest from its launch point and serves as a check to ensure our calculations align with physical reality.

Horizontal and vertical components

The horizontal and vertical components of motion are fundamental concepts in analyzing projectile motion. Separating the initial velocity into these two components allows us to understand different aspects of the motion independently.
The horizontal component, \(v_{0x}\), is calculated as \(v_0 \cos(\theta_0)\), and it stays constant throughout the projectile's flight because there's no horizontal acceleration.
The vertical component, \(v_{0y}\), is \(v_0 \sin(\theta_0)\). It changes over time due to gravity, which applies a vertical acceleration of \(-9.81 \mathrm{~m/s^2}\).
By focusing on these individual components, we can solve for key metrics such as time of flight, maximum height, and horizontal range. When analyzing the burrito's flight, knowing these components helps to precisely determine the point when the burrito is furthest horizontally and vertically from its starting position, thus maximizing the position vector's magnitude.