Problem 56 - 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 golfer hits a golf ball from the ground at an angle of \(60^{\circ}\) with respect to the ground at a velocity of 150 feet per second. (PICTURE CANNOT COPY)

Step-by-Step Solution

Answer

(a) Parametric equations: \(x(t) = 75t\), \(y(t) = 75\sqrt{3}t - 16t^2\). (b) Rectangular equation: \(y = \sqrt{3}x - \frac{16}{5625}x^2\). (c) Time in flight: 8.13 seconds, horizontal distance: 609.75 feet.

Step 1: Identify Initial Velocity Components

Given that the initial velocity is 150 feet per second and the launch angle is \(60^{\circ}\), we first determine the horizontal and vertical components of the velocity. These are given by \(v_{0x} = v_0 \cos \theta\) and \(v_{0y} = v_0 \sin \theta\), respectively. Thus, \(v_{0x} = 150 \cos(60^{\circ}) = 75\) feet per second and \(v_{0y} = 150 \sin(60^{\circ}) = 75\sqrt{3}\) feet per second.

Step 2: Derive Parametric Equations

The parametric equations for the projectile can be written using the initial velocity components. The horizontal motion is given by \(x(t) = v_{0x}t = 75t\). The vertical motion is influenced by gravity, so \(y(t) = v_{0y}t - \frac{1}{2}gt^2 = 75\sqrt{3}t - 16t^2\), where \(g = 32\) feet per second squared is the gravitational acceleration.

Step 3: Convert to Rectangular Equation

To find the rectangular equation, solve the horizontal parametric equation for \(t\): \(t = \frac{x}{75}\). Substitute this into the vertical parametric equation: \[ y = 75\sqrt{3}\left(\frac{x}{75}\right) - 16\left(\frac{x}{75}\right)^2 = \sqrt{3}x - \frac{16}{5625}x^2 \].

Step 4: Determine Time of Flight

The projectile is in flight until it returns to the ground, which is when \(y(t) = 0\). Solving the equation \(0 = 75\sqrt{3}t - 16t^2\) gives \(t(75\sqrt{3} - 16t) = 0\). The non-zero solution for \(t\) is \(t = \frac{75\sqrt{3}}{16}\), which simplifies to approximately 8.13 seconds.

Step 5: Calculate Horizontal Distance Covered

Use the time of flight to find the horizontal distance: \(x(8.13) = 75 \times 8.13 = 609.75\) feet.

Key Concepts

Parametric Equations

When dealing with projectile motion, parametric equations play a crucial role in describing the path that a projectile takes through space over time. The parametric equations stem from breaking down the motion into horizontal and vertical components.

Given a projectile launched with an initial velocity, it's split into horizontal and vertical components using trigonometric functions. For a launch angle of \(\theta\), the horizontal component is \(v_{0x} = v_0 \cos \theta\) and the vertical component is \(v_{0y} = v_0 \sin \theta\).

Using these, the motion equations can be formed:

• Horizontal motion: \(x(t) = v_{0x}t\)
• Vertical motion: \(y(t) = v_{0y}t - \frac{1}{2}gt^2\)

The horizontal motion shows constant velocity since there are no forces acting on it horizontally. The vertical motion, however, includes an acceleration term due to gravity (\(g = 32\) feet per second squared), reflected as a downward acceleration.

Rectangular Equation

A rectangular equation in projectile motion helps in viewing the path of a projectile as a single, unified path rather than through time intervals. To convert the parametric equations to a rectangular equation, solve the horizontal component for time \(t\) and replace it in the vertical equation.

From the parametric form, we have \(t = \frac{x}{75}\). Substituting this in for \(t\) in \(y(t)\) yields:

• \(y = 75\sqrt{3}\left(\frac{x}{75}\right) - 16\left(\frac{x}{75}\right)^2\)
• Simplify further to: \(y = \sqrt{3}x - \frac{16}{5625}x^2\)

This equation shows the path as a parabola, which is typical for projectile motion without air resistance. It makes it easier to visualize the trajectory on a coordinate plane.

Time of Flight

The time of flight in projectile motion is the total time that a projectile remains airborne, from launch until it lands back on the ground. To calculate this, set the vertical motion equation \(y(t)\) to zero since the projectile returns to ground level.

Rearranging the expression \(0 = 75\sqrt{3}t - 16t^2\), we find:

• Factor out \(t\): \(t(75\sqrt{3} - 16t) = 0\)
• This yields \(t = 0\) or \(t = \frac{75\sqrt{3}}{16}\)

The root \(t = 0\) represents the start of the motion, so the non-zero solution, approximately 8.13 seconds, is the total time of flight. Understanding the time of flight helps determine when and where the projectile will land.

Horizontal Distance

Horizontal distance in projectile motion represents how far the projectile travels horizontally during its entire flight. Once the time of flight is known, calculating horizontal distance becomes straightforward using the horizontal motion equation.

The horizontal distance \(x\) covered can be calculated by multiplying the horizontal velocity \(v_{0x}\) with the time of flight \(t\):

• \(x = v_{0x} \times t\)
• Substituting \(v_{0x} = 75\) ft/s and \(t \approx 8.13\) s: \(x = 75 \times 8.13 = 609.75\) feet

This result shows how far from the original launch point the projectile will land, which in this case, is approximately 609.75 feet. Knowing the horizontal distance is particularly useful in planning where the projectile should target.