During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.11 km/s at an initial inclination of 39.4◦ to the horizontal. The acceleration of gravity is 9.8 m/s2 . How far away did the shell hit? Answer in units of km.
To determine how far away the shell hit, we can use the equations of projectile motion.
The horizontal motion and vertical motion are independent of each other. In the horizontal direction, the initial velocity is the same throughout, while in the vertical direction, there is an acceleration due to gravity.
First, let's analyze the vertical motion of the shell. We need to determine the time it takes for the shell to reach the highest point of its trajectory and then come back down to the ground.
The initial vertical velocity, v₀y, is given by:
v₀y = v₀ * sin(θ)
Where:
v₀ = initial speed of the shell = 2.11 km/s = 2110 m/s
θ = initial inclination angle = 39.4°
Using these values, we can calculate v₀y:
v₀y = 2110 m/s * sin(39.4°)
Next, we'll calculate the time taken for the shell to reach its highest point. At the highest point, the vertical velocity will be zero.
The vertical displacement, Δy, at the highest point is given by:
Δy = v₀y * tₚ - (1/2) * g * tₚ²
Where:
tₚ = time to reach the highest point
g = acceleration due to gravity = 9.8 m/s²
Since the vertical velocity at the highest point is zero, we can rearrange the equation as follows:
0 = v₀y - g * tₚ
v₀y = g * tₚ
Solving for tₚ, we have:
tₚ = v₀y / g
Substituting the values, we get:
tₚ = (2110 m/s * sin(39.4°)) / 9.8 m/s²
Now that we have the time taken to reach the highest point, we can determine the total time of flight, t.
Since the shell takes the same amount of time to reach the highest point as it does to fall back down, the total time of flight is:
t = 2 * tₚ
Once we know the total time of flight, we can calculate the horizontal distance, d, using the horizontal velocity and time of flight.
The horizontal displacement, d, is given by:
d = v₀x * t
Where:
v₀x = initial horizontal velocity
To find v₀x, we can use the initial speed and the initial inclination angle:
v₀x = v₀ * cos(θ)
Substituting the values, we have:
v₀x = 2110 m/s * cos(39.4°)
Finally, we can calculate d:
d = (2110 m/s * cos(39.4°)) * 2 * tₚ
Substituting the calculated value of tₚ, we can find d:
d = (2110 m/s * cos(39.4°)) * 2 * (2110 m/s * sin(39.4°)) / 9.8 m/s²
Evaluating this expression will give us the distance the shell hit.
To find the distance at which the shell hit, we need to use the projectile motion equations.
The horizontal motion of the shell is unaffected by gravity, so we can use the equation:
Distance = (initial speed) x (time)
The vertical motion of the shell is affected by gravity, so we can use the equation:
Vertical distance = (initial vertical speed) x (time) + (1/2) x (acceleration due to gravity) x (time^2)
We can solve for the time of flight using the vertical motion equation.
Since the shell was fired at an initial inclination of 39.4 degrees to the horizontal, we can split the initial speed of 2.11 km/s into horizontal and vertical components.
Using trigonometry:
Initial horizontal speed = Initial speed x cos(initial inclination)
Initial vertical speed = Initial speed x sin(initial inclination)
Initial horizontal speed = 2.11 km/s x cos(39.4 degrees)
Initial vertical speed = 2.11 km/s x sin(39.4 degrees)
Now, we can find the time of flight:
Vertical distance = 0 (since the shell lands at the same height it was fired from)
0 = (2.11 km/s x sin(39.4 degrees)) x t + (1/2) x (9.8 m/s^2) x t^2
Solving this equation will give us the time of flight, t.
Once we have the time of flight, we can substitute it into the horizontal motion equation to find the distance at which the shell hit:
Distance = (2.11 km/s x cos(39.4 degrees)) x t
Now, let's calculate the distance.