Problem 44 - A Graphical Approach to College Algebra

Question

AProjectile If a projectile is fired at an angle of \(30^{\circ}\) with the horizontal, the parametric equations that describe its motion are $$x=v_{0} \frac{\sqrt{3}}{2} t, y=\frac{v_{0}}{2} t-16 t^{2}, \quad \text { for } t \text { in }[0, \infty)$$

Step-by-Step Solution

Answer

The projectile hits the ground at \(t = \frac{v_0}{32}\), traveling \(\frac{v_0^2 \sqrt{3}}{64}\) horizontally.

Step 1: Understand the Parametric Equations

The motion of the projectile is described by two parametric equations: \(x = v_0 \frac{\sqrt{3}}{2} t\) for the horizontal position, and \(y = \frac{v_0}{2} t - 16t^2\) for the vertical position. Here, \(v_0\) is the initial velocity and \(t\) is time.

Step 2: Find the Horizontal Range

Set the vertical equation \(y = \frac{v_0}{2} t - 16t^2\) to zero to find the time when the projectile hits the ground (since \(y=0\)).\[\frac{v_0}{2} t - 16t^2 = 0 \t(\frac{v_0}{2} - 16t) = 0 \t = 0 \text{ or } \frac{v_0}{2} = 16t \t = \frac{v_0}{32}\]The non-zero solution \(t = \frac{v_0}{32}\) gives us the total time of flight until the projectile hits the ground.

Step 3: Calculate the Total Horizontal Distance

To find the horizontal distance traveled, substitute \(t = \frac{v_0}{32}\) into the horizontal position equation \(x = v_0 \frac{\sqrt{3}}{2} t\).\[x = v_0 \frac{\sqrt{3}}{2} \left(\frac{v_0}{32}\right) = \frac{v_0^2 \sqrt{3}}{64}\]This equation represents the total horizontal distance the projectile will cover.

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The motion involves two components: horizontal motion and vertical motion, which together form a curved trajectory known as a parabola.

• The horizontal motion is uniform, meaning it has a constant velocity.
• The vertical motion is subjected to gravitational acceleration, which influences the upward and downward movement.

In the given problem, the projectile is launched at an angle of 30 degrees with an initial velocity \(v_0\). The parametric equations breaking down this motion are: - \(x(t) = v_0 \frac{\sqrt{3}}{2} t\) governing the horizontal position.- \(y(t) = \frac{v_0}{2} t - 16 t^2\) describing the vertical position.These equations allow us to track the projectile’s path over time, offering insights into its position, velocity, and the interaction between horizontal and vertical components.

Horizontal Range

The horizontal range of a projectile is the total horizontal distance it travels while in motion. To determine this, we focus on where the projectile lands. By solving the vertical position equation \(y(t) = \frac{v_0}{2} t - 16 t^2 = 0\) for when the projectile hits the ground \(y=0\), we find the total time of flight, which is \(t = \frac{v_0}{32}\). This means, from launch until the moment it strikes the ground again, the projectile remains airborne for this duration. Once we have the time of flight, it is substituted into the horizontal position equation \(x(t) = v_0 \frac{\sqrt{3}}{2} t\) to find the horizontal range: \[x = \frac{v_0^2 \sqrt{3}}{64}\] This formula shows that the range depends on the initial velocity squared, multiplied by a constant factor, displaying the interplay between speed and distance for the projectile.

Initial Velocity

Initial velocity is a crucial factor in determining a projectile's motion, including its range, peak height, and time of flight. It is the speed at which an object is launched and can be divided into horizontal and vertical components based on the launch angle. For this problem, the initial velocity is expressed in terms of the vector components:

• Horizontal: \(v_{0x} = v_0 \cos(30^\circ) = v_0 \frac{\sqrt{3}}{2}\)
• Vertical: \(v_{0y} = v_0 \sin(30^\circ) = \frac{v_0}{2}\)

These components are plugged into the parametric equations to describe the projectile's path. The horizontal component remains constant, whereas the vertical is influenced by gravity, causing the projectile to rise and then fall. Thus, understanding these components helps predict the overall trajectory.

Time of Flight

The time of flight for a projectile is the duration from when it is launched until it lands back on the ground. In calculating this, finding the time at which the vertical position \(y(t)\) returns to zero is necessary since it signifies the projectile has landed.In our case, solving the equation \( \frac{v_0}{2} t - 16 t^2 = 0 \) gives us two solutions: 0, at the launch, and \( t = \frac{v_0}{32} \), when it lands. We discard the zero since it represents the start time, thus indicating that \( t = \frac{v_0}{32} \) is the time of flight.Time of flight is dependent solely on the vertical motion of the projectile and gravitational forces. By knowing it, we can further calculate how far the projectile travels and analyze the trajectory path effectively.