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 1.66 km/s at

an initial inclination of 38.8 to the horizontal. The acceleration of gravity is 9.8 m/s^2

How far away did the shell hit? (km)
How long was it in the air? (s)

Well, if Big Bertha was shelling Paris, I hope she didn't hit any pastry shops. Those delicate macarons would have been in grave danger!

To calculate the distance the shell hit, we need to know the time of flight and horizontal component of velocity.

We can start by finding the horizontal component of velocity using trigonometry. The initial speed, 1.66 km/s, makes a 38.8° angle with the horizontal. So, the horizontal component of velocity is:

Vx = (1.66 km/s) * cos(38.8°)

And the time of flight can be found using the equation:

t = [2 * V * sin(θ)] / g

where V is the initial vertical component of velocity and θ is the initial inclination angle.

Now, let me plug in the numbers and calculate the funny answers for you! Oops, I mean the actual answers:

Vx = (1.66 km/s) * cos(38.8°)
t = [2 * (1.66 km/s) * sin(38.8°)] / 9.8 m/s^2

Calculating these values should give you the actual distance the shell hit and the time they spent in the air.

To solve this problem, we can use the equations of projectile motion. The horizontal motion and the vertical motion of the shell are independent of each other.

First, let's calculate the time of flight (total time the shell was in the air). We can use the vertical motion equation:

h = v0 * sin(θ) * t + (1/2) * g * t^2

where:
h = vertical displacement (0 since the shell returns to the same level)
v0 = initial vertical velocity = 1.66 km/s * sin(38.8°) = 1.01 km/s
θ = initial inclination = 38.8°
g = acceleration due to gravity = 9.8 m/s^2

Plugging these values into the equation:
0 = (1.01 km/s) * t - (1/2) * (9.8 m/s^2) * t^2

Simplifying:
0 = 1010 m/s * t - 4.9 m/s^2 * t^2

This equation is a quadratic equation. To find the time of flight, we can solve for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac))/(2a)

where:
a = -4.9 m/s^2
b = 1010 m/s
c = 0

Plugging these values into the quadratic formula:
t = (-1010 ± √(1010^2 - 4(-4.9)(0)))/(2 * -4.9)

Simplifying:
t ≈ (-1010 ± √(1020100))/(2 * -4.9)

Taking the positive root gives us the time of flight:
t ≈ (1010 + √(1020100))/(2 * 4.9)
t ≈ 206.45 seconds

Now, let's calculate the horizontal distance traveled by the shell. We can use the horizontal motion equation:

d = v0 * cos(θ) * t

where:
d = horizontal distance traveled
v0 = initial horizontal velocity = 1.66 km/s * cos(38.8°) = 1.32 km/s
θ = initial inclination = 38.8°
t = time of flight = 206.45 seconds

Plugging these values into the equation:
d = (1.32 km/s) * (206.45 seconds)
d ≈ 272.06 km

Therefore, the shell hit a distance of approximately 272.06 km away, and it was in the air for approximately 206.45 seconds.

To find the distance the shell hit and the time it was in the air, we can use the principles of projectile motion.

1. First, let's break down the initial velocity of the shell into its horizontal and vertical components.

The horizontal component (Vx) remains constant throughout the motion because there is no horizontal acceleration. Therefore, Vx = initial velocity * cos(initial inclination)
Vx = 1.66 km/s * cos(38.8°)

The vertical component (Vy) is affected by gravity and changes over time. Therefore, Vy = initial velocity * sin(initial inclination)
Vy = 1.66 km/s * sin(38.8°)

2. Next, let's analyze the vertical motion to find the time of flight (t).

The equation that relates vertical displacement (h), initial vertical velocity (Vy), time of flight (t), and acceleration due to gravity (g) is:
h = Vy * t + (1/2) * g * t^2

Since the shell was launched horizontally, the initial vertical displacement (h) is zero. We can rewrite the equation as:
0 = Vy * t + (1/2) * g * t^2

Simplifying the equation:
(1/2) * g * t^2 = - Vy * t

3. From the equation above, we can solve for the time of flight (t).

(1/2) * g * t^2 + Vy * t = 0

We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = (1/2) * g, b = Vy, and c = 0

Substituting the values:
t = (-Vy ± √(Vy^2 - 4 * (1/2) * g * 0)) / (2 * (1/2) * g)

Simplifying the equation:
t = (-Vy ± √(Vy^2)) / g
t = (-Vy ± Vy) / g

Since the time of flight cannot be negative, t = (2 * Vy) / g

4. Now, let's find the horizontal distance (R) traveled by the shell.

The horizontal distance traveled is given by the equation:
R = Vx * t

Substituting the values:
R = (1.66 km/s * cos(38.8°)) * ((2 * (1.66 km/s * sin(38.8°)) / 9.8 m/s²))

Simplifying the equation:
R = (1.66 * 1000 m/s * cos(38.8°)) * ((2 * (1.66 * 1000 m/s * sin(38.8°)) / 9.8 m/s²))

Calculating both distance and time will give us the answers to the questions.

Let's plug in the values and calculate the distance and time of flight.