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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

An airplane is traveling from New York to Paris, a distance of $5.80 \times 10^{3} \mathrm{km} .$ Ignore the curvature of the Earth. (a) If the cruising speed of the airplane is \(350.0 \mathrm{km} / \mathrm{h},\) how much time will it take for the airplane to make the round-trip on a calm day? (b) If a steady wind blows from New York to Paris at \(60.0 \mathrm{km} / \mathrm{h},\) how much time will the round-trip take? (c) How much time will it take if there is a crosswind of \(60.0 \mathrm{km} / \mathrm{h} ?\)
An arrow is shot into the air at an angle of \(60.0^{\circ}\) above the horizontal with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) (a) What are the \(x\) - and \(y\) -components of the velocity of the arrow \(3.0 \mathrm{s}\) after it leaves the bowstring? (b) What are the \(x\) - and \(y\) -components of the displacement of the arrow during the 3.0 -s interval?
A gull is flying horizontally \(8.00 \mathrm{m}\) above the ground at $6.00 \mathrm{m} / \mathrm{s} .$ The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, (a) what is the horizontal distance to the rocks at the moment that the gull should let go of the clam? (b) With what speed relative to the rocks does the clam smash into the rocks? (c) With what speed relative to the gull does the clam smash into the rocks?
Sheena can row a boat at \(3.00 \mathrm{mi} / \mathrm{h}\) in still water. She needs to cross a river that is 1.20 mi wide with a current flowing at $1.60 \mathrm{mi} / \mathrm{h} .$ Not having her calculator ready, she guesses that to go straight across, she should head \(60.0^{\circ}\) upstream. (a) What is her speed with respect to the starting point on the bank? (b) How long does it take her to cross the river? (c) How far upstream or downstream from her starting point will she reach the opposite bank? (d) In order to go straight across, what angle upstream should she have headed?
A pilot starting from Athens, New York, wishes to fly to Sparta, New York, which is \(320 \mathrm{km}\) from Athens in the direction $20.0^{\circ} \mathrm{N}\( of \)\mathrm{E} .$ The pilot heads directly for Sparta and flies at an airspeed of \(160 \mathrm{km} / \mathrm{h}\). After flying for $2.0 \mathrm{h}$, the pilot expects to be at Sparta, but instead he finds himself \(20 \mathrm{km}\) due west of Sparta. He has forgotten to correct for the wind. (a) What is the velocity of the plane relative to the air? (b) Find the velocity (magnitude and direction) of the plane relative to the ground. (c) Find the wind speed and direction.
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