# Extremely hard physics problem! Help!

1. Jan 10, 2005

### BlasterV

The German army in World War I shelled Paris with very large artillery pieces. These so-called 'Paris Guns' were actually a combination of two mortars stuck together. Pointed 55 degrees above the horizon, they fired shells with an initial speed of 1700 meters per second.

The guns were first used on March 23, 1918, a few days after the German army launched its last great offensive on the Western front. Three of the guns were located in Crepy-en-Laonnoise (about 8 km west-north-west of Laon), just behind the German lines.

Ok, I will show the parts I have solved so far first.

In the absence of air resistance, what was the range of the 'Paris' guns?

D = 277113.43613 m

How long was the shell in the air?

Time = 284.195607202 s

What is the distance from Crepy-en-Laonnoise to Paris?

D = 100 km (aka 100000 meters)

These 2 questions below are where it all falls apart:

If the Germans had wanted to shell London from the same spot in Crepy, what muzzle velocity would the shells have needed?

The distance from London to Crepy is 382797.19 m

Placing artillery on high ground extends its range. If the Germans wanted to shell London from Crepy with their existing 'Paris' guns, to what altitude would they have needed to lift the guns?

To be honest, all my attempts have been incorrect and I don't even think I'm close, so if someone can show me how these 2 parts are done it'd help alot.

I couldn't even comeup with a system to solve the first problem, so I have no work to show.

As for the second: (And none of this may even be remotely close) I tried it this way:

Vo x = 975.079941797
D Crepy to London/ Vo = T for mortar to make trip
T = 392.580314281 s

D = Vo y * t - 4.9 ( t^2 )
Answer is positive, since its height above ground.
D = 208493.541605 m

Anyway, need help really bad, thanks!

Additional info: Vo y to paris = 1392.55847574 m/s

2. Jan 10, 2005

### HallsofIvy

Staff Emeritus
Presumably you know the formulas:

x= x0+ vxt and
y= y0+ vyt- (g/2)t2

where x and y are the horizontal and vertical (distance and height) coordinates of the shell at time t. vx and vy are the x and y components of the orignal velocity and g is the acceleration due to gravity. I imagine you used those formulas to solve the first problems.

The first of the two problems does not give v0 so leave that as an unknown. Of course, T is also unknow but you still have two equations to solve for those two unknowns. Since the distance from Crepy to London is 382797.19 m, you have to solve (taking x0 and y0 both 0)
x= vxt= 382797.19 and
y= vyt- (g/2)t2= 0

Assuming we are still using the 55 degree angle, vx= v cos(55) and vy= v sin(55) where v is the unknown muzzle velocity.

In order to solve the second problem, use the same v as initially given but now include y0, the unkown initial height. Solve
x= 1700cos(55)t= 382797.19 and
y= y0+ 1700 sin(55)t- (g/2)t2 for t and y0.

3. Jan 10, 2005

### Curious3141

I think these are fairly simple particle mechanics, but the answer (especially to the second part) seems unrealistic.

Basically if the ammo starts at the origin of a Cartesian coordinate system, the coordinates of the ammo at time t $(x_t, y_t)$ can be represented by :

$$x_t = (v\cos\theta)t$$
$$y_t = (v\sin\theta)t - \frac{1}{2}gt^2$$

For the first problem :

"If the Germans had wanted to shell London from the same spot in Crepy, what muzzle velocity would the shells have needed?

The distance from London to Crepy is 382797.19 m"

We need to put $x_t = D = 382797.19 m$ and $y_t = 0$ and solve for $v$.

You will get an expression like :

$$v = \sqrt{\frac{gD}{\sin2\theta}}$$

which you can then evaluate. You should get $v = 1998 m/s$

For the second part :

"Placing artillery on high ground extends its range. If the Germans wanted to shell London from Crepy with their existing 'Paris' guns, to what altitude would they have needed to lift the guns?"

I'm assuming the muzzle velocity is 1700 m/s and the ground distance from London to Crepy is as already given. Let $h$ be the required elevation in meters.

Here you need $v = 1700 m/s, x_t = D = 382797.19 m$ and $y_t = -h$

Work through the equations and you should come to :

$$h = D(\frac{gD}{2v^2\cos^2\theta} - \tan\theta)$$

Plug in the values to determine $h$. I wasn't happy with this one because I get a value like 208493.5 m, which is over 200 km altitude ! This isn't realistic as far as I can see, but the math should be right.

4. Jan 10, 2005

### BlasterV

I followed the part of getting v = 1998 m/s and it was right, but before you tried to help me I already had 208493.5m which is wrong :( is there something small we are both missing?

5. Jan 10, 2005

### Andrew Mason

I assume the 55 angle fixed. It seems odd because the maximum range for level ground is at 45 degrees (I think it is less for an elevated launch).

For the second part, I think all you have to plug $t = x_{Paris}/v_0cos\theta$ into the equation of motion: $y = v_0sin\theta t - \frac{1}{2}gt^2$ to find y (it will be negative). $v_0 = 1700 m/sec.$ so it should be easy enough.

I won't give you the answer. In real life I would give you the wrong one just to be sure you missed Paris.

AM

6. Jan 10, 2005

### Curious3141

The problem is I did exactly this but got the same answer as the orig. poster, which he says is wrong,

7. Jan 11, 2005

### Andrew Mason

Why is it wrong? Try some different launch angles. The horizontal speed (0 launch angle) is 1700 m/sec so it is going to take a minimum of 225 seconds to get there. An object falls a long way in 225 seconds (about 250 km with no air resistance). Raising the angle will give it more 'hang time' but y is still going to be on the order of 100 km. This is a rather poor question.

Here is another way of looking at it: The maximum height occurs at time: t=142.1 seconds after launch ($v_y=0$ when $v_{y0}=gt)$. In this time it moves 138 km (range 276 km with level target). ($x = v_{0x}t$). In order to reach London it it has to travel for an additional 108 seconds after it passes the level target distance (ie. when it is going down at 1392m/sec it drops for another 108 seconds). How far do you think that shell will drop in 108 seconds?: a very long distance!

AM

Last edited: Jan 11, 2005
8. Jan 11, 2005

### Curious3141

I think we can all agree this is a dumb question. :yuck: The only "difficulty" is in the unnecessary tedium of dragging out the problem as a long and pseudo-romantic "true" war story for which the numbers need to be ridiculously unwieldy and difficult to work with. And after all that trouble, the final answer is not even remotely realistic. :rofl:

There are a lot of better ways to teach particle mechanics than setting tedious problems with big numbers.

Last edited: Jan 11, 2005
9. Jan 12, 2005

### BigStelly

The mathematics to this one most of you did seems to make sense, the reason they used an angle higher than 45degrees, is because density in air varies as you go up, it gets thinner thereby lowering air resistance and allowing the projectile to go further. It was a strange phenomenon not encountered till the early 20th century because guns were not yet powerful enough to reach a height to where this would make a difference.

But yes I would say the numbers you have all come up with make sense, that is also what i got.