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 39.4β¦ to the horizontal. The acceleration of gravity is 9.8 m/s2 . How far away did the shell hit? Answer in units of km.
Range = (Vo^2/g)*sin(2A)
A = 39.4 degrees
Vo = 2110 m/s
See http://www.phy.ilstu.edu/~holland/phy220/Comp_Prob_3.pdf
Plug in the numbers and crank out the answer. Then convert range in meters to km
Range = 446 km
The actual muzzle veloclity and range were quite a bit less than that, according to the reference.
To find the distance that the shell hit, we can use the equations of projectile motion. The horizontal displacement of a projectile can be calculated using the formula:
π₯ = π£βπ‘ cos(ΞΈ)
Where:
- π₯ is the horizontal displacement
- π£β is the initial speed of the projectile
- π‘ is the time of flight
- ΞΈ is the angle of inclination
First, we need to convert the initial speed from km/s to m/s:
2.11 km/s * (1000 m/1 km) = 2110 m/s
Next, let's find the time of flight. In order to do this, we need to consider the vertical motion of the shell. The vertical displacement can be determined using the equation of motion:
π¦ = π£βπ‘ sin(ΞΈ) β Β½ππ‘Β²
Where:
- π¦ is the vertical displacement
- π£β is the initial speed of the projectile
- π‘ is the time of flight
- ΞΈ is the angle of inclination
- π is the acceleration due to gravity
Since the vertical displacement is zero when the shell hits the ground, we can set π¦ = 0 and solve for π‘:
0 = (2110 m/s)(π‘ sin(39.4Β°)) - Β½(9.8 m/sΒ²)(π‘Β²)
0 = 2110(π‘ sin(39.4Β°)) - 4.9(π‘Β²)
0 = π‘(2110 sin(39.4Β°) - 4.9π‘)
Solving this quadratic equation will give us the time of flight π‘. Plugging it back into the horizontal displacement equation will give us the final answer.