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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 m long, and it had a muzzle speed of 2.20 km/s. When the gun’s angle of elevation was set to 55.0°, what would be the range? For the purposes of solving this problem, ignore air resistance. (The actual range at this elevation was 121 km; air resistance cannot be ignored for the high muzzle speed of the shells.)

User Loman
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2 Answers

3 votes

Final answer:

The question requires solving a problem related to projectile motion within the historical context of World War I artillery, specifically referencing 'Big Bertha'. It involves calculating the range of the gun using the initial muzzle velocity and angle of elevation, assuming no air resistance.

Step-by-step explanation:

The question deals with the scenario of a cannon fire during war, specifically during World War I, involving the assessment of projectile motion without air resistance. The historical context introduces the 'Big Bertha,' a notable artillery piece. For the physics problem, assuming no air resistance, the projectile motion can be solved by breaking the velocity into horizontal and vertical components and then using kinematic equations.

Big Bertha's range when fired at an elevation of 55.0° with a muzzle speed of 2.20 km/s can be calculated using the formula for projectile motion: Range = (v^2 × sin(2 × angle)) / g, where v is the muzzle velocity, angle is the angle of elevation, and g is the acceleration due to gravity.

In general, projectile motion problems require understanding of the following concepts:

Initial velocity

Maximum height

Range of the projectile

Earth's curvature effect on projectile motion

User NGR
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4.3k points
4 votes

Answer:

464.1 km

Step-by-step explanation:

Using R = u²sin2Ф/g where R = range of projectile, u = muzzle speed = 2.20 km/s = 2.20 × 10³ m/s, Ф = angle of elevation or projection angle = 55° and g = acceleration du° to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

R = u²sin2Ф/g

R = (2.20 × 10³ m/s)²sin2(55°)/9.8 m/s²

R = 4.84 × 10⁶ m²/s²sin110°/9.8 m/s²

R = 4.84 × 10⁶ m²/s² × 0.9397/9.8 m/s²

R = 4.548 × 10⁶ m²/s²/9.8 m/s²

R = 0.4641 × 10⁶ m

R = 464.1 × 10³ m

R = 464.1 km

User Craig Graham
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