Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Do the following. (a) Determine the parametric equations that model the path of the projectile. (b) Determine the rectangular equation that models the path of the projectile. (c) Determine the time the projectile is in flight and the horizontal distance covered. A model rocket is launched from the ground with a velocity of 48 feet per second at an angle of \(60^{\circ}\) with respect to the ground.

Step-by-Step Solution

Answer

(a) Parametric: \(x(t) = 24t\), \(y(t) = 24\sqrt{3}t - 16t^2\). (b) Rectangular: \(y = \sqrt{3}x - \frac{x^2}{36}\). (c) Flight time: \(\frac{3\sqrt{3}}{2}\) sec, Distance: \(36\sqrt{3}\) feet.

Step 1: Break Down Initial Velocity

The initial velocity of 48 feet per second at an angle of \(60^{\circ}\) can be resolved into horizontal and vertical components. The horizontal component \(v_{0x}\) is given by \(v_{0x} = 48 \cos(60^{\circ})\), and the vertical component \(v_{0y}\) is given by \(v_{0y} = 48 \sin(60^{\circ})\). Calculating these gives \(v_{0x} = 24\) feet per second and \(v_{0y} = 24\sqrt{3}\) feet per second.

Step 2: Write Parametric Equations

The parametric equations use the initial velocity components and time \(t\). The horizontal position \(x(t)\) is \(x(t) = v_{0x} \cdot t = 24t\). The vertical position \(y(t)\) is given by the equation \(y(t) = v_{0y} \cdot t - \frac{1}{2}gt^2 = 24\sqrt{3}t - 16t^2\) where \(g = 32 \text{ ft/s}^2\).

Step 3: Convert to Rectangular Equation

To find the rectangular equation, eliminate \(t\) from the parametric equations. From \(x(t) = 24t\), we get \(t = \frac{x}{24}\). Substitute this into \(y(t)\): \(y = 24\sqrt{3}\times \frac{x}{24} - 16\left(\frac{x}{24}\right)^2\). Simplifying gives \(y = \sqrt{3}x - \frac{x^2}{36}\).

Step 4: Determine Time of Flight

The projectile reaches the ground when \(y(t) = 0\). Solve \(24\sqrt{3}t - 16t^2 = 0\) for \(t\). Factor to get \(t(24\sqrt{3} - 16t) = 0\). This gives \(t = 0\) (launch time) and \(t = \frac{24\sqrt{3}}{16} = \frac{3\sqrt{3}}{2}\). The time of flight is \(t = \frac{3\sqrt{3}}{2}\) seconds.

Step 5: Find Horizontal Distance Covered

Since the projectile lands when \(t = \frac{3\sqrt{3}}{2}\), substitute this time into the horizontal equation to find distance: \(x = 24 \times \frac{3\sqrt{3}}{2} = 36\sqrt{3}\). The horizontal distance covered is \(36\sqrt{3}\) feet.

Key Concepts

Projectile Motion

Projectile motion is a type of motion where an object is thrown near the earth's surface, moving along a curved path under the action of gravity only. The motion can be broken into two components:

• Horizontal motion at a constant velocity.
• Vertical motion affected by gravitational acceleration.

For a model rocket launched with an initial velocity, such as 48 feet per second, the trajectory it follows can be analyzed by breaking the launch speed into horizontal and vertical components. The launch angle, like the given 60 degrees, greatly influences these components:

• The horizontal component, calculated by multiplying the initial velocity by the cosine of the angle.
• The vertical component, calculated by multiplying by the sine of the angle.

These components allow us to write parametric equations representing the position as functions of time. The resulting path shows a parabolic arc, typical of projectile motion.

Rectangular Equation

The rectangular equation provides a relation between the horizontal and vertical displacements by eliminating the time variable from parametric equations. It's a convenient way to study the trajectory shape without concerning ourselves directly with time.When analyzing our model rocket, we start with the parametric equations derived from initial velocity components. By expressing time from the horizontal equation, we can substitute this into the vertical equation.For example, with the horizontal position equation, like:\[ x(t) = 24t \]we solve for time:\[ t = \frac{x}{24} \]Substituting this into the vertical equation allows us to obtain a formula in terms of x and y only:\[ y = \sqrt{3}x - \frac{x^2}{36} \]This equation reaffirms the path of the rocket as a parabola, which represents the fundamental nature of projectile motion without time dependency.

Time of Flight

In projectile motion problems, the time of flight is the total time the projectile remains in the air. It is determined by solving for when the height or vertical position becomes zero, returning to the launch level.For our model rocket, given the vertical equation:\[ y(t) = 24\sqrt{3}t - 16t^2 \]set to zero to find the time after launch when it lands. Solving:\[ 24\sqrt{3}t - 16t^2 = 0 \]factoring leads to separate solutions:\[ t = 0 \quad \text{or} \quad t = \frac{3\sqrt{3}}{2} \]The non-zero solution is our total time of flight. This indicates how long the rocket travels before touching the ground again. In our model's case, it remains airborne for \( \frac{3\sqrt{3}}{2} \) seconds.

Horizontal Distance

The horizontal distance covered by a projectile is crucial as it demonstrates the range of its motion. This distance is computed once we determine the time of flight, by substituting it back into the horizontal position function.With an example equation:\[ x(t) = 24t \]on finding the total time of flight as \( \frac{3\sqrt{3}}{2} \) seconds, substitute to obtain:\[ x = 24 \times \frac{3\sqrt{3}}{2} \]This results in a straightforward calculation yielding the total horizontal range:\[ x = 36\sqrt{3} \] feet,indicating the rocket's maximum reach from launch in a straight line horizontally. This distance is directly influenced by both the initial velocity and the angle of launch, essential for predicting landing points of projectiles.