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 2.39 km/s at
an initial inclination of 49.6

to the horizontal.
The acceleration of gravity is 9.8 m/s
2
.
How far away did the shell hit?
Answer in units of km

To determine the distance at which the shell hits, we can use the projectile motion equations.

First, let's convert the initial velocity to m/s:

Initial velocity (v₀) = 2.39 km/s = 2.39 * 1000 m/s = 2390 m/s

Next, we need to find the time of flight, which is the total time it takes for the shell to reach the ground. The time of flight can be determined using the equation:

Time of flight (t) = (2 * initial velocity * sin(initial inclination angle)) / gravity

t = (2 * 2390 * sin(49.6°)) / 9.8

Now, let's calculate the time of flight:

t = (2 * 2390 * sin(49.6°)) / 9.8 ≈ 394.13 s

Finally, we can calculate the horizontal distance using the equation:

Horizontal distance (d) = initial velocity * cos(initial inclination angle) * time of flight

d = 2390 * cos(49.6°) * 394.13

Simplifying further:

d ≈ 579,613.4 m

Converting the distance to kilometers:

d ≈ 579.6134 km

Therefore, the shell hit a distance of approximately 579.6134 km away.

To find out how far away the shell hit, we can use the equations of projectile motion. We need to first calculate the time it took for the shell to hit the ground. Then, we can find the horizontal distance traveled by multiplying the time by the initial horizontal velocity.

Given:
Initial speed of the shell (u) = 2.39 km/s
Initial inclination (angle) = 49.6 degrees to the horizontal
Acceleration due to gravity (g) = 9.8 m/s^2

Step 1: Convert the initial speed to m/s
Since the given acceleration due to gravity is in m/s^2, it is better to convert the initial speed to m/s as well. 1 km = 1000 m, so the initial speed is:
u = 2.39 km/s × 1000 m/km
u = 2390 m/s

Step 2: Resolve the initial velocity into horizontal and vertical components.
The horizontal component (ux) can be found using the equation:
ux = u × cos(angle)
ux = 2390 m/s × cos(49.6°)

The vertical component (uy) can be found using the equation:
uy = u × sin(angle)
uy = 2390 m/s × sin(49.6°)

Step 3: Calculate the time of flight.
The vertical motion of the shell follows the equation:
y = uy × t - (1/2) × g × t^2
where y is the vertical displacement, uy is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time of flight.
The shell hits the ground when y = 0, so we can solve for t.

0 = uy × t - (1/2) × g × t^2
(1/2) × g × t^2 = uy × t
(1/2) × g × t = uy
t = (2 × uy) / g

Step 4: Calculate the horizontal distance.
The horizontal distance, denoted as D, traveled by the shell can be found using the equation:
D = u × t
where u is the initial speed and t is the time of flight calculated in the previous step.

D = u × t
D = 2390 m/s × t

Step 5: Substitute the values into the equations and calculate.
Substitute the values we have in each step and calculate the final result.

Solving Step 2:
ux = 2390 m/s × cos(49.6°) ≈ 1546.5 m/s
uy = 2390 m/s × sin(49.6°) ≈ 1772.3 m/s

Solving Step 3:
t = (2 × 1772.3 m/s) / 9.8 m/s^2 ≈ 362.85 s

Solving Step 4:
D = 1546.5 m/s × 362.85 s ≈ 561,440.025 m

Step 6: Convert the distance to kilometers.
D = 561,440.025 m × (1 km / 1000 m)
D ≈ 561.44 km

Therefore, the shell hit the ground approximately 561.44 km away.

vertical problem

Vi = 2390 sin 49.6

v = Vi - 9.8 t
at top v = 0
so
t = Vi/9.8 at top
that is half the trip
hits ground at tg = Vi/4.9

Now the horizontal problem
u = 2390 sin 49.6
range = u tg
range = 2390 sin 49.6 * Vi/4.9
so
range = 2390^2 sin 49.6 cos49.6 /4.9
= 575369 meters
= 575 km