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Chapter 2: Problem 15

A parachute-manufacturing company uses the formula \(d=3.69 \sqrt{\frac{m}{v^{2}}}\) to model the diameter, \(d,\) in metres, of its dome- shaped circular parachutes so that an object with mass, \(m,\) in kilograms, has a descent velocity, \(v,\) in metres per second, under the parachute. a) What is the landing velocity for a \(20-\mathrm{kg}\) object using a parachute that is \(3.2 \mathrm{m}\) in diameter? Express your answer to the nearest metre per second. b) A velocity of \(2 \mathrm{m} / \mathrm{s}\) is considered safe for a parachutist to land. If the parachute has a diameter of \(16 \mathrm{m}\) what is the maximum mass of the parachutist, in kilograms?

Short Answer

Expert verified
a) 5 m/s. b) 75 kg.

Step by step solution

01

Identify known values for part (a)

For part (a), identify the given values: the mass of the object is 20 kg and the diameter of the parachute is 3.2 meters.
02

Substitute known values into the formula for part (a)

Using the formula for the diameter, substitute the given values to find the descent velocity, \[ d=3.69 \sqrt{\frac{m}{v^{2}}}\] Plug in the values: \[ 3.2=3.69 \sqrt{\frac{20}{v^{2}}}\]
03

Solve for the descent velocity (v) for part (a)

Isolate \(v\) by performing algebraic operations. First, divide both sides by 3.69: \[\frac{3.2}{3.69} = \sqrt{\frac{20}{v^2}}\] Square both sides to get rid of the square root: \[\left(\frac{3.2}{3.69}\right)^2 = \frac{20}{v^2}\] Calculate the left side: \[\left(0.8672\right)^2 \approx 0.752 \] Solving for \(v^2\): \[0.752 = \frac{20}{v^2}\] Rearrange to solve for \(v\): \[ v^{2} = \frac{20}{0.752} \] \[ v \approx \sqrt{26.6} \approx 5.16 \] Round to the nearest metre per second: \[ \approx 5 \text{ m/s} \]
04

Identify known values for part (b)

For part (b), identify the given values: the velocity of descent is 2 m/s and the diameter of the parachute is 16 meters.
05

Substitute known values into the formula for part (b)

Using the formula for the diameter, substitute the given values to find the maximum mass, \[ d=3.69 \sqrt{\frac{m}{v^{2}}}\] Plug in the values: \[ 16 = 3.69 \sqrt{\frac{m}{2^{2}}}\]
06

Solve for the maximum mass (m) for part (b)

Isolate \(m\) by performing algebraic operations. First, divide both sides by 3.69: \[ \frac{16}{3.69} = \sqrt{\frac{m}{4}}\] Square both sides to get rid of the square root: \[ \left(\frac{16}{3.69}\right)^2 = \frac{m}{4} \] Calculate the left side: \[ \left( 4.336 \right)^2 \approx 18.798 \] Solving for \(m\): \[ 18.798 \cdot 4 = m \rightarrow m \approx 75.191 \text{ kg}\] The maximum mass rounded to the nearest kilogram is: \[ \approx 75 \text{ kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Descent Velocity
In parachute descent calculations, the descent velocity ( v ) plays a crucial role. It's the speed at which the object or person descends with the parachute. The formula given in the exercise is designed to relate the diameter of the parachute, its mass, and this velocity. For a specific mass and parachute diameter, you can calculate the descent velocity.

This relationship is essential for ensuring safety. Too high a descent velocity could result in injury upon landing. Conversely, a very low descent velocity might be impractical for the length of the descent. To find the descent velocity, you often need to rearrange the provided formula and solve for v . This involves algebraic manipulations, including dividing by constants, squaring both sides, and taking square roots.

By performing these operations correctly, you can accurately determine the descent velocity for different masses and parachute diameters. This ensures that safety standards are met and that descent speeds are manageable.
Mass and Diameter Relationship
The mass of the object (or person) using the parachute and the diameter of the parachute ( d ) have a direct relationship, defined by the given formula. The formula shows that larger parachutes can support more significant masses at safer or lower descent velocities.

This relationship is crucial when designing parachutes for different weights. For heavier objects, the parachute diameter needs to be larger to ensure a safe descent velocity. Conversely, for lighter objects, smaller parachutes suffice.

To understand this relationship better, consider the formula: d = 3.69 . When solving for an unknown d or m , observing how mass and diameter interact helps explain why certain parachute sizes are chosen for specific weights. This understanding is vital when making practical decisions about parachute equipment.
Algebraic Manipulations
Algebraic manipulations are a cornerstone of solving parachute descent problems. They involve a series of steps to isolate the variable of interest, which could be the descent velocity ( v ) or the mass ( m ).

For example, if you need to find the descent velocity, you set up the equation with known mass and diameter values, then perform operations like dividing both sides by constants, squaring to eliminate square roots, and taking the square root of both sides.

Similarly, when determining the mass an object can have for a given diameter and descent velocity, you follow steps to isolate m . This can involve reversing the steps used to find the velocity, carefully applying each algebraic operation.

The key to mastering these problems is a methodical approach: identify all known values, substitute them into the formula, and follow through with the necessary algebraic steps to isolate and solve for the unknown variable. Consistent practice of these operations builds confidence and accuracy in solving similar problems in the future.

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