How far away did the shell hit?

Question

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 2.35 km/s at an initial inclination of 41.7 ◦ 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.

How long was it in the air? Answer in units of s.

Answer 1

First, we will assume that this is on flat ground (no hills or earth curvature) and there is no drag.

we know its x position is given by

x(t) = v_x*t

where v_x is initial velocity in the x-direction (caused by Big Bertha firing). Because we're not considering the force of drag, our x(t) has no forces acting on it, causing no acceleration. v_x is found by considering the velocity from the cannon and the angle of its fire

v_x = v*Cos(θ) = 2.35*Cos(41.7º) km/s

so our final x(t) equation is given by

***x(t) = (2.35*Cos(41.7º) km/s)*t *** (note this is in km)

once we know the time it crashes, we can plug in that into x(t_crash) to find the total distance traversed.

Our y-position is given by

y(t) = v_y*t -g*t²/2 = (2.35*Sin(41.7º) km/s)*t - 9.8m/s²*t²/2

where v_y is the initial velocity in the y-direction (Sin instead of Cos). Note that the left part of the equation is in km while the right part is in meters. So lets change our equation slightly.

y(t) = (2.35*Sin(41.7º) km/s)*t - .0098km/s²*t²/2

To find t_crash, we must find the times when y(t) = 0.

y(t_crash) = (2.35*Sin(41.7º) km/s)*t_crash - .0098km/s²*t²_crash/2 = 0

--> t_crash[(2.35*Sin(41.7º) km/s) - .0098km/s²*t_crash/2] = 0

--> 2.35*Sin(41.7º) km/s - .0098km/s²*t_crash/2 = 0

==> t_crash =2*( 2.35*Sin(41.7º) km/s)/(.0098km/s²)

This gives you the time-of-flight (or the time the shell impacts) in seconds. Plug that answer into our ***x(t)*** equation to find the total horizontal distance traversed!

Answer 2

Hi Jimmy,

To answer 2D kinematics problems like this, we must first use first try to find time (t), as this variable is used in both the x and y directions. As gravity works in the vertical direction, we must start by analyzing variables in the y direction. We are given

1. Initial velocity: v = 2.35 km/s = 2350 m/s
2. Inclination: 41.7°
3. Acceleration due to gravity: a = -9.8 m/s² (negative sign to account for gravity acting downwards)

We can find the Initial vertical velocity: v_0y = v * sin(θ) = 2350 m/s * sin(41.7°) = 1563 m/s.

The final vertical velocity is the same as the initial vertical velocity, but coming down instead of going up. This is written as v_y = -v_0y = -1563 m/s (negative to show that the object is coming down).

The kinematic equation that includes v_y, v_0y, acceleration a, and time t is

v_y = v_0y + at ⇒ t = (v_y - v_0y) / a

We can substitute our values into this equation to solve for t, how long the shell was in the air:

t = (-1563 m/s -1563 m/s) / (9.8 m/s²) = 319 s

To find how far away the shell hit, we use displacement = (horizontal velocity) * time, or

Δx = v_xt,

The horizontal velocity is v_x = v * cos(θ) = 2350 m/s * cos(41.7°) = 1755 m/s, and we found time to be t = 319 s. Plugging these values into the displacement equation gives us

Δx = (1755 m/s)(319 s) = 559845 m, or about 560 km.

Hope this helps answer your question! Feel free to reach out if you have any questions.

Akshat Y.