Question

Ask you to solve a variety of problems based on the principles of projectile motion. A Cessna flies at \(1000 f t\) at \(150 m p h\) and drops a box of supplies to the professor (and his wife) on an island. Ignoring wind resistance, how far horizontally will the supplies travel before they land?

Short answer

The supplies will travel approximately 1738 feet horizontally.

Step-by-step solution

Convert the Speed to Feet per Second

First, we need to convert the speed of the Cessna from miles per hour to feet per second. The conversion factors are: 1 mile = 5280 feet and 1 hour = 3600 seconds. So, the speed is \(150 \text{ mph} \times \frac{5280 \text{ ft}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3600 \text{ seconds}} = 220 \text{ ft/s}\).

Calculate the Time of Flight

Next, we determine how long the box is in the air. The box is dropped from a height of 1000 feet and only gravity affects its vertical motion. The formula to use is \(d = \frac{1}{2}gt^2\), where \(d\) is 1000 feet, \(g\) is 32.2 feet per second squared (acceleration due to gravity), and \(t\) is the time in seconds. Solving \(1000 = \frac{1}{2} \times 32.2 \times t^2\), we find \(t = \sqrt{\frac{1000}{16.1}} \approx 7.9 \text{ seconds}\).

Calculate the Horizontal Distance

Finally, use the time of flight to determine the horizontal distance traveled. The horizontal speed is 220 feet per second and the time is approximately 7.9 seconds. The horizontal distance \(x\) is given by \(x = velocity \times time = 220 \times 7.9 = 1738 \text{ feet} \).

Conclusion

Therefore, the supplies will travel approximately 1738 feet horizontally before they land.

Key concepts

Horizontal Distance Calculation

To calculate the horizontal distance an object travels in projectile motion, we must consider its horizontal velocity and the time it remains in motion. In this case, the Cessna travels at 150 mph, but first, we need to establish its velocity in feet per second. Using the conversion factors of 1 mile = 5280 feet and 1 hour = 3600 seconds, the speed is converted to 220 feet per second.

Once we have the horizontal velocity, we need the time of flight, which we'll discuss in another section. The horizontal distance is then found by multiplying the horizontal velocity by this time. This gives the distance the object will travel before landing.

Gravity Effects

Gravity plays a crucial role in projectile motion as it influences the vertical motion of any object in free fall. In this problem, we ignore wind resistance, so the only force acting on the box after it's dropped is gravity. Gravity causes the box to accelerate downwards at a constant rate, which is approximately 32.2 feet per second squared on Earth.

The vertical motion can be described using the formula \( d = \frac{1}{2} g t^2 \), where \( d \) represents the vertical distance (1000 feet in this exercise), \( g \) is the gravitational acceleration, and \( t \) is the time in seconds. Solving this allows us to find the time it takes for the box to land, in this case, roughly 7.9 seconds.

Unit Conversion

Understanding unit conversion is essential for solving physics problems, especially in projectile motion scenarios where different units are often used. Here, the airplane's speed needed to be converted from miles per hour (mph) to feet per second (ft/s).

To convert, multiply the speed by the number of feet in a mile (5280 feet) and then divide by the number of seconds in an hour (3600 seconds). This conversion changes the speed of the plane from 150 mph to 220 ft/s, making further calculations consistent and manageable.

Physics Problem Solving

Solving physics problems involving projectile motion requires a structured approach that includes understanding the problem, identifying known values, selecting appropriate formulas, and careful calculations.

• Firstly, clearly define the problem: In this scenario, it's the horizontal distance supplies travel.
• Utilize unit conversion right away to work with suitable measurements (miles per hour to feet per second).
• Apply gravity effects to discover how long the object remains in motion.
• Finally, use the correct formula to solve for the horizontal distance.

This systematic method ensures all aspects of the problem are addressed, leading to a concise and accurate solution.