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Chapter 3: Problem 9

A bullet weighing 1 oz is fired vertically downward from a stationary helicopter with a muzzle velocity of \(1200 \mathrm{ft} / \mathrm{sec}\). The air resistance (in pounds) is numerically equal to \(16^{-5} v^{2}\), where \(v\) is the velocity (in feet per second). Find the velocity of the bullet as a function of the time.

Short Answer

Expert verified
The velocity of the bullet as a function of time is given by: $$ v(t) = \frac{1}{\sqrt{32.2} \cdot 16^{-\frac{5}{2}}} \sin\left(\sqrt{32.2} t\right) $$

Step by step solution

01

Write down the given parameters and adjust units.

We are given: - The bullet's weight: \(1 oz\). The weight force should be in pounds (\(lb\)) so we need to divide it by 16. Thus, \(W = \frac{1}{16} \, lb\). - Muzzle velocity: \(v_0 = 1200 \, ft/s\) (Given in the problem) - Air resistance: \(R(v) = 16^{-5}v^2 \, lb\), where \(v\) is the velocity in \(ft/s\).
02

Write the governing equation.

Now, considering the forces acting on the bullet, we get the equation: $$ m\frac{dv}{dt} = W - R(v) $$ Here, \(m\) is the mass of the bullet and \(\frac{dv}{dt}\) is the time rate of change of velocity (acceleration).
03

Convert weight into mass

To convert the weight of the bullet (in pounds) to mass (in slugs), we use the definition of weight: $$ W = mg $$ where \(g = 32.2 \, ft/s^2\) is the acceleration due to gravity. So, we get: $$ m = \frac{W}{g} = \frac{1}{16} \times \frac{1}{32.2} \, slug $$
04

Substitute parameters into the governing equation.

Now, let's substitute the parameters (gravity, mass and air resistance) into the governing equation: $$ \frac{1}{16} \times \frac{1}{32.2} \frac{dv}{dt} = \frac{1}{16} - 16^{-5} v^2 $$
05

Separate variables and integrate.

We first separate the variables and then integrate both sides to find the velocity of the bullet as a function of time: $$ \frac{dv}{1 - (32.2 \times 16^{-5} v^2)} = dt $$ Integrating both sides: $$ \int_0^v \frac{1}{1 - (32.2 \times 16^{-5} v^2)} dv = \int_0^t dt $$ Using substitution \(u = 16^{-\frac{5}{2}} \sqrt{32.2} v\), we get: $$ \frac{1}{\sqrt{32.2}} \int_0^u \frac{1}{1 - u^2} du = t $$ Now integrate the left side: $$ \frac{1}{\sqrt{32.2}} \left[\arcsin(u)\right]_0^u = t $$
06

Write the final expression for velocity as a function of time.

Now, we can write the expression for the velocity of the bullet as a function of time: $$ v(t) = \frac{1}{\sqrt{32.2} \cdot 16^{-\frac{5}{2}}} \sin\left(\sqrt{32.2} t\right) $$ This is the velocity of the bullet as a function of time.

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Most popular questions from this chapter

A girl on her sled has just slid down a hill onto a level field of ice and is starting to slow down. At the instant when their speed is \(5 \mathrm{ft} / \mathrm{sec}\), the girl's father runs up and begins to push the sled forward, exerting a constant force of \(15 \mathrm{lb}\) in the direction of motion. The combined weight of the girl and the sled is \(96 \mathrm{lb}\), the air resistance (in pounds) is numerically equal to one-half the velocity (in feet per second), and the coefficient of friction of the runners on the ice is \(0.05\). How fast is the sled moving \(10 \mathrm{sec}\) after the father begins pushing?

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