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 49°, what would be the range? For the purposes of solving this problem, neglect air resistance.
Range = 2(V^2/g)sin49cos49
= (V^2/g)sin98 = 2.21*10^5 m = 221 km
To solve this problem, we can use the equations of motion for projectile motion. The horizontal motion and vertical motion are independent of each other, so we can analyze them separately.
First, let's find the time of flight (t) for the projectile. We'll use the vertical (y) motion equation:
y = v₀y * t + (1/2) * a * t²
Since the shell is fired at an angle of 49°, the initial vertical velocity (v₀y) is calculated by:
v₀y = v₀ * sin(θ)
Here, v₀ is the initial velocity (muzzle speed) and θ is the angle of elevation.
In this case, the initial velocity v₀ is 1.48 km/s (1480 m/s), and the angle of elevation θ is 49°. So, we can calculate v₀y as follows:
v₀y = 1480 * sin(49°)
Next, let's find the time of flight (t). For the shell to reach its maximum height, its vertical velocity will become zero. Therefore:
v₀y = v₀ * sin(θ) = 0
Since acceleration in the vertical direction (a) is due to gravity and is -9.8 m/s², we can solve for t as follows:
0 = v₀y - 9.8 * t
Now, let's solve this equation for t:
t = v₀y / 9.8
Substituting the value of v₀y, we get:
t = (1480 * sin(49°)) / 9.8
Now that we have calculated the time of flight (t), let's find the horizontal distance traveled (range). We'll use the horizontal (x) motion equation:
x = v₀x * t
Since there is no acceleration in the horizontal direction, the initial horizontal velocity (v₀x) remains constant throughout the projectile's motion. The initial horizontal velocity can be calculated by:
v₀x = v₀ * cos(θ)
Here, v₀ is the initial velocity (muzzle speed) and θ is the angle of elevation.
In this case, the initial velocity v₀ is 1.48 km/s (1480 m/s), and the angle of elevation θ is 49°. So, we can calculate v₀x as follows:
v₀x = 1480 * cos(49°)
Finally, substituting the values of v₀x and t into the equation for range (x), we can calculate the range as follows:
range = v₀x * t
Let's do the calculations:
v₀y = 1480 * sin(49°) = 1138.98 m/s
t = (1480 * sin(49°)) / 9.8 = 116.12 s
v₀x = 1480 * cos(49°) = 948.19 m/s
range = 948.19 * 116.12 = 110116.33 m
Therefore, when the gun's angle of elevation was set to 49°, the range of the shells fired from Big Bertha would be approximately 110.12 km.
To calculate the range of the long-range gun known as Big Bertha, we can use the equations of projectile motion. The range is the horizontal distance traveled by the shell when it lands on the ground.
Here's how we can solve the problem step by step:
1. Convert the muzzle speed to meters per second:
The muzzle speed of Big Bertha is given as 1.48 km/s. We need to convert it to m/s by multiplying it by 1000 (since there are 1000 meters in a kilometer).
1.48 km/s * 1000 = 1480 m/s
2. Calculate the horizontal component of the velocity:
The horizontal component of the velocity (Vx) remains constant throughout the projectile's flight, assuming no air resistance. We can calculate it by multiplying the muzzle speed by the cosine of the angle of elevation.
Vx = 1480 m/s * cos(49°)
3. Calculate the vertical component of the velocity:
The vertical component of the velocity (Vy) changes due to the acceleration due to gravity. We can calculate it by multiplying the muzzle speed by the sine of the angle of elevation.
Vy = 1480 m/s * sin(49°)
4. Calculate the time of flight:
The time of flight (t) is the total time that the shell is in the air. Since the vertical motion is purely due to gravity, we can calculate the time of flight using the vertical component of velocity as follows:
t = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
5. Calculate the range:
The range (R) is the horizontal distance traveled by the shell. We can calculate it by multiplying the horizontal component of velocity by the time of flight.
R = Vx * t
Now, let's calculate the range using the given values:
Angle of elevation: 49°
Muzzle speed: 1.48 km/s (1480 m/s)
Acceleration due to gravity: 9.8 m/s^2
Vx = 1480 m/s * cos(49°)
Vy = 1480 m/s * sin(49°)
t = (2 * Vy) / g
R = Vx * t
By substituting the values into the equations, we can calculate the range.