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 1.48 km/s. When the gun's angle of elevation was set to 50°, what would be the range? (The actual range at this elevation was 121 km; air resistance cannot be neglected for the high muzzle speed of the shells.)

To find the range of the gun, we can break down the problem into two components: The horizontal range and the vertical range. We'll start by finding the time of flight using the vertical motion equation. Then, we'll use the horizontal motion equation to calculate the horizontal range.

Step 1: Finding the time of flight
We can use the vertical motion equation:
y = V₀yt - (1/2)gt²

Where:
V₀y is the initial vertical velocity (in this case, the muzzle velocity in the vertical direction)
g is the acceleration due to gravity (approximately 9.8 m/s²)
y is the vertical displacement (in this case, the vertical distance is 0, since the shell lands at the same height it was fired from)
t is the time of flight

Rearranging the equation, we get:
t = 2V₀y / g

Step 2: Finding the horizontal range
We can use the horizontal motion equation:
x = V₀xt

Where:
V₀x is the initial horizontal velocity (in this case, the muzzle velocity in the horizontal direction)
t is the time of flight
x is the horizontal displacement (the range we want to find)

We can break down V₀ into horizontal and vertical components:
V₀x = V₀ × cos(θ)
V₀y = V₀ × sin(θ)

Where:
θ is the angle of elevation in degrees

Substituting the values, we get:
x = (V₀ × cos(θ)) × (2V₀y / g)

Step 3: Calculate the range
Now we can substitute the given values into the equation:
V₀ = 1.48 km/s (convert to m/s)
θ = 50°
g = 9.8 m/s²

First, convert the muzzle velocity to m/s:
V₀ = 1.48 km/s = 1.48 × 1000 m/s = 1480 m/s

Using the equation above:
x = (1480 m/s × cos(50°)) × (2 × (1480 m/s × sin(50°)) / 9.8 m/s²)

Simplifying the equation:
x = (1480 m/s × cos(50°)) × (2 × (1480 m/s × sin(50°)) / 9.8 m/s²)
x ≈ 132,213.8 m

So, at an angle of elevation of 50°, the range of the gun would be approximately 132,213.8 meters or 132.2 kilometers. However, this does not account for air resistance, so the actual range is different (given to be 121 km in the question). The difference in range is due to the effect of air resistance on the shell's trajectory.