Problem 82 - Precalculus with Limits

Question

PROJECTILE MOTION In Exercises 81 and 82, consider the path of a projectile projected horizontally with a velocity of \(v\) feet per second at a height of \(s\) feet, where the model for the path is \(x^2 = -\dfrac{v^2}{16}(y-2)\). In this model (in which air resistance is disregarded), \(y\) is the height (in feet) of the projectile and \(x\) is the horizontal distance (in feet) the projectile travels. A cargo plane is flying at an altitude of 30,000 feet and a speed of 540 miles per hour. A supply crate is dropped from the plane. How many \(\textit{feet}\) will the crate travel horizontally before it hits the ground?

Step-by-Step Solution

Answer

The supply crate will travel approximately 152,843 feet horizontally before it hits the ground.

Step 1: Convert Initial Speed to Feet per Second

The speed of the plane is given in miles per hour. For solving this exercise, we have to convert it to feet per second since the projectile's model uses these units. 1 mile is 5280 feet and 1 hour is 3600 seconds. Therefore, the velocity \(v\) is \(v = 540 \times \dfrac{5280}{3600} = 792 \) feet per second.

Step 2: Use Given Altitude and Speed in the Equation

The equation \(x^2 = -\dfrac{v^2}{16}(y-2)\) represents the path of the crate. Insert the altitude \(s=30,000\) feet into \(y\) and velocity \(v=792\) feet per second to solve for \(x\)

Step 3: Solve Equation for x

After substituting these values into the equation, we get \(x^2 = -\dfrac{(792)^2}{16}(30,000-2)\). To solve this for \(x\), take the square root of both side to get \(x = \sqrt{-\dfrac{(792)^2}{16}(30,000-2)}\). Calculate the value of \(x\) which represents the horizontal distance the crate will travel before it hits the ground.

Key Concepts

Velocity Conversion

When working on projectile motion problems, the units of velocity are crucial for ensuring accurate calculations. In this exercise, the velocity of the cargo plane is initially provided in miles per hour (mph). To align with the projectile motion model's units, which use feet and seconds, we need a conversion.
To convert from mph to feet per second (fps), first remember:

• 1 mile = 5280 feet
• 1 hour = 3600 seconds

Thus, for the velocity conversion: 1. Multiply the speed in miles per hour by the number of feet in a mile: 540 mph x 5280 = 2,851,200 feet per hour 2. Then, convert feet per hour to feet per second by dividing by 3600 seconds: 2,851,200 / 3600 = 792 fps Hence, the plane's speed is 792 feet per second.

Horizontal Distance Calculation

Once velocity is converted, solving for horizontal distance involves substituting known values into the projectile motion equation. This equation is given by: \[ x^2 = -\dfrac{v^2}{16}(y-2) \] Where:

• \(x\) is the horizontal distance in feet.
• \(v\) is the converted velocity, 792 feet per second.
• \(y\) is the initial height, 30,000 feet.

Substitute these values into the equation:\[ x^2 = -\dfrac{792^2}{16}(30,000-2) \]After performing the calculations:

• Calculate \( v^2 = 792^2 = 627,264 \)
• Substitute into the equation: \( x^2 = -\dfrac{627,264}{16}(29,998) \)

Finally, solve for \(x\) by taking the square root:\[ x = \sqrt{ -\dfrac{627,264}{16} \times 29,998 } \]This calculation gives the horizontal distance traveled by the crate.

Mathematical Modeling

Mathematical models provide us a way to simulate real-world scenarios using equations, fostering understanding through simplification. In projectile motion, models help predict a projectile's path based on its initial conditions like speed and height.
This specific model, \( x^2 = -\dfrac{v^2}{16}(y-2) \), is derived from kinematic equations for motion under constant acceleration, reflecting gravity's role. It's tailored for scenarios where air resistance is ignored, which simplifies the calculations and focuses on idealized motion.
Key elements of this model include:

• Initial Velocity (\(v\)): Determines how fast and far the object flies horizontally.
• Initial Height (\(y\)): Influences the time the projectile is airborne.
• Gravity: Implied in the constant \( -\dfrac{1}{16} \), governing the object's trajectory.

By substituting values into this model, as shown, we get the projectile's path in an ideal world free of air resistance.

Neglecting Air Resistance

In the realm of projectile motion modeling, simplifying assumptions often include disregarding air resistance. This assumption allows the use of straightforward equations, such as the one applied in this exercise, to describe the path of a projectile.
Neglecting air resistance means:

• Assuming the only force acting on the projectile is gravity.
• Simplifying trajectory calculations, making them more manageable as no drag force is accounted for.

While this assumption introduces some differences from real-life scenarios, it is highly effective for educational purposes. It helps students understand fundamental motion physics without additional complexities. In scenarios where precision isn't critical, or when just beginning to explore projectile motion, this simplification makes learning more approachable.