Mechanics: Simple Projectile Motion – 2 (Fort and Ship)

Here is a projectile problem that fascinated me, and I’ve been meaning to show it to you. It comes from Neville de Mestre, “The Mathematics of Projectiles in Sport”, 1990. This is the second post about simple projectile motion, so you might want to look at the first one.

Here we go.

A fort is on top of a cliff h meters directly above the ocean. Approaching the fort is a ship whose guns have the same muzzle velocity vo as the guns at the fort….

Find over what range the ship can be fired on, from the fort, without being able to effectively return the fire.

If gh is small compared with vo^2 show that this distance is approximately double the height of the cliff.

So, we need to find two distances: max from fort to ship, and max from ship to fort.

Here are the underlying equations:

(As before, sx and sy are the x- and y-components of position s; vx and vy are the x- and y-components of velocity v; sxo and syo are the initial values of position components sx and sy; vxo and vyo are the initial values of velocity components vx and vy; -g is the acceleration of gravity; t is time.)

My boundary values are chosen so that the projectile starts at the origin:

And for this problem, we know the muzzle velocity vo, so we need to rewrite the x and y components of the initial velocity in terms of vo:

Introducing the boundary conditions, the underlying equations become

That is, the first two equations, for position, are:

Now, at some unknown time T the projectile will be at a height h, so we set t = T and sy(T) = h; we’re just renaming the two things:

We solve the second equation for T, getting two answers… then choose the positive (larger) value…

Plug into e1 and find the maximum… that is, find the angle \alpha\ that maximizes the horizontal distance sx at time T. We do that by setting the first derivative to zero, and solving for \alpha\ .

(You may recall that for h = 0, i.e. when the target is at the same height as the launch point, maximum range is obtained when \alpha\ = 45{}^{\circ}. But our target is not at the same height as the gun.)

Having put that comment between the description and the screenshot, let me repeat the description: the horizontal distance is… then we substitute our formula for the time T… and then we set the derivative wrt \alpha\ to zero to find where the horizontal distance is an extremum (and on physical grounds I’m sure it’s a maximum):

Now we actually solve for the angle \alpha\ at which the derivative is zero; Mathematica® gives us four solutions:

The hassle is picking out the right solution (\alpha\ in the first quadrant). That should be the positive arc cos of a positive number. Here, the problem is that I think vo^2 is the dominant term… ah, but it is subtracted in both numerator and denominator, so that should cancel.

I choose solution 4:

By plugging that into our equation for sx(T), we get a general answer:

OK, let’s check that with some numbers. Let me take a muzzle velocity of 400 meter/sec, and h = -100 meter (we’re computing for the fort, aiming at a ship at lower height).

Interesting. I was initially surprised that 45{}^{\circ} is effectively optimal, but… we wil see that compared to the horizontal range, the difference in heights is insignificant.

Here are the equations we got for sx(T) and sy(T), at maximum range… We solve the first one for T… and then look at the second one…

There are two solutions… the negative solution is for a projectile fired at the fort from below; it’s so close that we’re effectively on the linear portion of the cosine curve, shooting 100 meter up at 45{}^{\circ}, so we’re almost exactly 100 m to the left of the fort.

The positive solution says the fort can hit the ship when it’s 16415.2 m away. How many miles is that?

Convert[16415. Meter,Mile]

= 10.1998 Mile.

So we’re talking roughly a vertical difference of 300 feet over a horizontal range of 50,000 feet… 3 in 500 is 0.6%. That’s why the angles are so close to 45° for maximum range: the final altitudes are very close together compared to the range.

Now for the ship… muzzle velocity vo is the same, but now h is +100 m, positive instead of negative:

Now we put that angle into our equations for sx and sy at maximum range…

Once again we have two solutions. One is for hitting something on the way up, right after we fire, almost exactly 100 m away… but not exactly, because the angle isn’t exactly 45{}^{\circ}.

As for the long solution (on the way down)… it is almost exactly 200 m less than the range of the fort: the ship cannot return fire until it is 200 m closer than where the fort can hit it (16215.2 meter versus 16415.2).

Let’s look at that in general. Since I worked with arbitrary h. we have… change the sign of h…

and subtract the two right-hand-sides:

… and then let Mathematica find the limit as vo goes to infinity:

(I couldn’t very well let h -> 0 and expect an answer involving h, so instead I let \text{vo} \rightarrow \infty\ .)

Since h was 100 m, the limit is 200 m, and that’s what we got.

Now, this problem is rather extreme: the difference in heights is 100 meters, while the range is slightly more than 16,000 meters. On a drawing, we would be hard-pressed to see the differences in the trajectories, or the difference in altitudes.

So let me make some related drawings where the differences are visible. But I’m going to do that next week. For now I’ll just summarize the key fact: the projectile from the fort falls an additional 100 m below the fort — and travels an additional 100 m horizontally at the same time; meanwhile, the projectile from the ship must stop 100 m up, rather than falling to sea level — at the same time that projectile fails to travel 100 m horizontally. The combination is the 200 m difference, in the limit as the angle \alpha\ approaches 45°.


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