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 0.725 km/s at
an initial inclination of 73.7◦
to the horizontal.
The acceleration of gravity is 9.8 m/s2
How far away did the shell hit?
Answer in units of km.
How long was it in the air?
Answer in units of s.
Vi = 725 sin 73.7 = 725*0.96 = 696 m/s upward at start
m g H = (1/2) m Vi^2
H = 696^2 / (2 * 9.81) = 48422/(2*9.81) = 24,680 meters max height
average upward speed = 696/2 = 348 m/s
so time upward = 24,680/348 = 70.9 seconds upwards
total time = 2 * 70.9 = 142 seconds
constant horizontal speed = 725 cos 73.7 = 203.5 m/s
range = 203.5 * 142 = 28,895 m = 29 kilometers (yikes !)
Well, to calculate the distance the shell hit and how long it was in the air, we'd need to use some projectile motion equations. But before we start, let me just say that Big Bertha sounds like a really big lady with a powerful punch!
Now, let's get down to business. To find the distance the shell hit, we can use the horizontal motion equation:
distance = initial velocity * time
Since the shell was fired at an initial inclination of 73.7◦ to the horizontal, we need to find the horizontal component of its initial velocity. We can do that by multiplying the initial velocity by the cosine of the angle:
horizontal velocity = initial velocity * cos(angle)
Now we can plug in the numbers:
horizontal velocity = 0.725 km/s * cos(73.7◦)
Using a bit of trigonometry magic, we get:
horizontal velocity ≈ 0.725 km/s * 0.274
horizontal velocity ≈ 0.19865 km/s
Now, to find the time the shell was in the air, we can use the vertical motion equation:
time = 2 * (initial vertical velocity) / acceleration due to gravity
The initial vertical velocity can be found by multiplying the initial velocity by the sine of the angle:
vertical velocity = initial velocity * sin(angle)
So let's plug in the numbers:
vertical velocity = 0.725 km/s * sin(73.7◦)
Using more trigonometry magic, we get:
vertical velocity ≈ 0.725 km/s * 0.961
vertical velocity ≈ 0.69785 km/s
Now we can calculate the time:
time = 2 * (0.69785 km/s) / 9.8 m/s²
Uh-oh, we have a problem here. Our velocity is in kilometers per second, but our acceleration is in meters per second squared. We need to convert our velocity to meters per second before plugging in the numbers.
1 km/s = 1000 m/s
So, the vertical velocity in meters per second is:
vertical velocity = 0.69785 km/s * 1000 m/s per km
vertical velocity ≈ 697.85 m/s
Now we can calculate the time:
time = 2 * (697.85 m/s) / 9.8 m/s²
time ≈ 142.231 s
Finally, to find the distance the shell hit, we can use the horizontal velocity and the time:
distance = horizontal velocity * time
distance ≈ 0.19865 km/s * 142.231 s
distance ≈ 28.29565 km
So, the shell hit approximately 28.3 km away. Now that's quite a punch from Big Bertha!
To find the distance the shell hit, we can use the equation of motion:
d = (v^2 * sin(2θ)) / g
Where:
d = distance (what we need to find)
v = initial speed = 0.725 km/s
θ = initial inclination = 73.7 degrees = 73.7 * (π/180) radians
g = acceleration due to gravity = 9.8 m/s^2
First, let's convert the initial speed to m/s:
v = 0.725 km/s * 1000 m/km = 725 m/s
Now, let's convert the initial inclination to radians:
θ = 73.7 * (π/180) radians = 1.286 radians
Now we can substitute the values into the equation:
d = (725^2 * sin(2*1.286)) / 9.8
Calculating this equation:
d ≈ 51,558.68 m
Converting the distance to km:
d ≈ 51.56 km
So, the shell hit approximately 51.56 km away.
To find the time the shell was in the air, we can use the equation of motion:
t = (2 * v * sin(θ)) / g
Substituting the values:
t = (2 * 725 * sin(1.286)) / 9.8
Calculating this equation:
t ≈ 183.16 s
So, the shell was in the air for approximately 183.16 seconds.
To find how far away the shell hit, we need to determine its horizontal distance traveled. We can use the equation for horizontal distance:
𝑑 = 𝑣₀𝑥 × 𝑡
Where:
- 𝑑 is the horizontal distance
- 𝑣₀𝑥 is the initial horizontal velocity
- 𝑡 is the time of flight
First, let's convert the initial velocity to m/s:
𝑣₀ = 0.725 km/s = 725 m/s
Since the initial inclination is given with respect to the horizontal, we can find the initial horizontal velocity using trigonometry:
𝑣₀𝑥 = 𝑣₀ × cos(𝜃)
where 𝜃 = 73.7°
Converting the angle to radians:
𝜃_radians = 73.7° × (𝜋/180°)
Now, we can calculate 𝑣₀𝑥:
𝑣₀𝑥 = 725 m/s × cos(𝜃_radians)
Using this value, we can determine how far away the shell hit by calculating 𝑑.
To find the time of flight, we need to determine the total time it takes for the shell to hit the ground. For this, we can use the equation for vertical motion:
𝑑 = 𝑣₀𝑦 × 𝑡 + (1/2)𝑎𝑡²
where:
- 𝑑 is the vertical distance (which we assume to be the height of the target or the height of the projectile's path since gravity is a downward acceleration)
- 𝑣₀𝑦 is the initial vertical velocity (0 since the shell was fired horizontally)
- 𝑎 is the acceleration due to gravity (-9.8 m/s², negative because it acts downwards)
- 𝑡 is the time of flight
Rearranging the equation, we get:
𝑡 = -2𝑑/𝑎
Since we are interested in the time of flight for the entire trajectory, we need to use twice the vertical distance traveled. So, 𝑡 = -2(𝑑/2)/𝑎 simplifies to 𝑡 = -𝑑/𝑎.
By substituting 𝑑 for the height of the projectile's path (or the height of the target), we can find 𝑡.
Now, let's calculate the horizontal distance and the time of flight.