Question

Falling body When an object falling from rest encounters air resistance proportional to the square of its velocity, the distance it falls (in meters) after \(t\) seconds is given by \(d(t)=\frac{m}{k} \ln (\cosh (\sqrt{\frac{k g}{m}} t)),\) where \(m\) is the mass of the object in kilograms, \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(k\) is a physical constant. a. A BASE jumper \((m=75 \mathrm{kg})\) leaps from a tall cliff and performs a ten-second delay (she free-falls for \(10 \mathrm{s}\) and then opens her chute). How far does she fall in 10 s? Assume \(k=0.2\) b. How long does it take for her to fall the first 100 m? The second \(100 \mathrm{m} ?\) What is her average velocity over each of these intervals?

Short answer

Answer: The distance fallen by the BASE jumper after 10 seconds is approximately 249.05 meters. The time it takes for them to fall the first 100 meters is about 4.432 seconds, and the time for the second 100 meters is about 3.861 seconds. Their average velocities over the first and second 100 meters are approximately 22.563 m/s and 25.902 m/s, respectively.

Step-by-step solution

Find the distance fallen

To find the distance d(t) fallen by the BASE jumper, we need to substitute the given values into the formula provided: \( d(t) = \frac{m}{k} \ln(\cosh( \sqrt{ \frac{k g}{m} } t )) \) Given values: \(m=75 kg, t=10 s, k=0.2,\) and \(g=9.8 \frac{m}{s^2}\).

Substitute and calculate

Now, we can substitute the values into the formula: \( d(10) = \frac{75}{0.2} \ln( \cosh( \sqrt{ \frac{0.2 \cdot 9.8}{75} } \cdot 10 )) \) After we calculate, we'll get the distance fallen after 10 seconds: \( d(10) \approx 249.05m \) Part b:

Find the time for first 100 meters

To find the time it takes to fall the first 100 meters, we'll set d(t) equal to 100 meters and solve for t: \( 100 = \frac{75}{0.2} \ln(\cosh( \sqrt{ \frac{0.2 \cdot 9.8}{75} } t )) \) We can solve for t numerically. The result is: \( t \approx 4.432 s \)

Find the time for second 100 meters

To find the time it takes to fall the second 100 meters, we'll set d(t) equal to 200 meters and solve for t: \( 200 = \frac{75}{0.2} \ln(\cosh( \sqrt{ \frac{0.2 \cdot 9.8}{75} } t )) \) We can solve for t numerically. The result is: \( t \approx 8.293 s \) Now, we can subtract the first-time result from the second-time result to get the time interval for falling the second 100 meters: \(8.293s - 4.432s = 3.861s\)

Calculate average velocities

Now, we can calculate the average velocity for each interval. 1. First 100 meters: \( \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{4.432} \approx 22.563\frac{m}{s}\) 2. Second 100 meters: \( \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{3.861} \approx 25.902\frac{m}{s}\) Summary of results: - Distance fallen in 10 seconds: 249.05 meters - Time to fall first 100 meters: 4.432 seconds - Time to fall second 100 meters: 3.861 seconds - Average velocity over first 100 meters: 22.563 m/s - Average velocity over second 100 meters: 25.902 m/s

Key concepts

Air Resistance

When an object falls freely through the air, it doesn't just move under the force of gravity. It also encounters air resistance, which is a force opposing its motion. In many practical problems, air resistance is assumed to be proportional to the square of the object's velocity. This means that as the velocity of the object increases, the air resistance increases exponentially.

• Air resistance is dependent on factors such as the density of the air, the cross-sectional area of the object, and the drag coefficient.
• Unlike gravitational force, which is constant, air resistance increases with velocity, introducing a complex non-linear factor in calculations of motion.

This resistance reduces the acceleration of the object, causing it to eventually reach a terminal velocity, where the net force acting on it is zero. At this stage, the object falls at a uniform speed because the upward force of air resistance equals the downward force of gravity.

Distance Calculation

To calculate how far an object falls in a given time, we often incorporate both gravitational forces and air resistance into our equations. In the given exercise, the distance function is expressed as:\[ d(t) = \frac{m}{k} \ln(\cosh(\sqrt{\frac{k g}{m}} t)) \]

• Here, \(d(t)\) represents the distance fallen after time \(t\).
• \(m\) is the mass of the body, and \(k\) is a constant that depends on conditions like air density and shape of the object.

With this formula, you can compute the exact distance covered in a certain time, taking air resistance into account. This helps us understand not just how far, but also how the speed varies during the fall.

Average Velocity

Average velocity is a key concept in understanding the movement of objects over time. It is the total displacement divided by the total time taken. In the case of falling bodies with air resistance, it gives a clearer picture of the overall speed during specific intervals.
In the exercise:

• For the first 100 meters, the average velocity is calculated by dividing the distance (100 meters) by the time taken for that distance.
• Similarly, for the second 100 meters, the calculation involves a different time interval since the object's speed increases with fall duration.

The formula for average velocity is:\[ \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} \]Despite visible changes in speed due to air resistance, this calculation provides an important simplification and helps in comparing different segments of the fall.

Numerical Solution

Sometimes, solving equations analytically (with an exact formula) is complex, especially with non-linear equations involving \(\cosh\). Instead, numerical methods are often used to find approximate solutions. In this problem, numerical solutions were used to find the time required to fall certain distances.

• For instance, finding when the distance \(d(t)\) equals 100 meters involves solving a non-linear equation numerically.
• Such numerical techniques often use computer algorithms because they handle complex expressions and iterative calculations efficiently.

While exact solutions give precise answers, numerical approximations are crucial for problems that are difficult to solve analytically. They allow us to get close approximations that are often sufficiently accurate for practical purposes.