/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 The citizens of Paris were terri... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 84

The citizens of Paris were terrified during World War I when they were suddenly bombarded with shells fired from a long-range gun known as Big Bertha. The barrel of the gun was \(36.6 \mathrm{m}\) long and it had a muzzle speed of \(1.46 \mathrm{km} / \mathrm{s} .\) When the gun's angle of elevation was set to \(55^{\circ},\) what would be the range? For the purposes of solving this problem, neglect air resistance. (The actual range at this elevation was \(121 \mathrm{km} ;\) air resistance cannot be ignored for the high muzzle speed of the shells.)

Short Answer

Expert verified
Answer: The range of Big Bertha neglecting air resistance is approximately 203.73 km.

Step by step solution

01

Convert speed to m/s and find horizontal and vertical components of the initial velocity

First, we have to convert the given speed from km/s to m/s. To do this, we multiply the speed by 1000: \(1.46 \mathrm{km/s} * 1000 = 1460 \mathrm{m/s}\) Now, we'll find the horizontal and vertical components of the initial velocity using the angle of elevation. Remember that: \(v_{x} = v_{0} * \cos(\theta)\) and \(v_{y} = v_{0} * \sin(\theta)\) Given the angle of elevation \(\theta = 55^{\circ}\) and the initial speed \(v_{0} = 1460 \mathrm{m/s}\), we can find the horizontal and vertical components as follows: \(v_{x} = 1460 \mathrm{m/s} * \cos(55^{\circ}) = 834.30\mathrm{m/s}\) \(v_{y} = 1460 \mathrm{m/s} * \sin(55^{\circ}) = 1196.96\mathrm{m/s}\)
02

Find the time of flight

Now, we will find the time of flight of the projectile. This can be done using the vertical component of the velocity and the following equation: \(t = \frac{2 * v_{y}}{g}\) where g is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). \(t = \frac{2 * 1196.96 \mathrm{m/s}}{9.81\mathrm{m/s^2}} = 244.16\mathrm{s}\)
03

Find the range of the projectile

Finally, we can find the range of the projectile using the horizontal component of the velocity and the time of flight: \(R = v_{x} * t\) \(R = 834.30\mathrm{m/s} * 244.16\mathrm{s} = 203734.17\mathrm{m}\) Since we want our answer in kilometers, we'll divide by 1000: \(R \approx 203.73\mathrm{km}\) Hence, the range of Big Bertha neglecting air resistance would be approximately \(203.73\mathrm{km}\).

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Most popular questions from this chapter

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