Simple Projectile Motion 3 – Between roof and ground

Let’s try a much shorter range problem. The ship and the fort in the previous post were shooting at each other from 10 miles apart, and the ship could not return fire for about a 220 yard interval.

Reduce the muzzle velocity to 20 meter/second; change h to -10 meter for the high ground firing at the low ground. With these numbers, this situation is more like two guys throwing rocks at each other; one of them is on the roof of a 2-story building and the other is at street level.

Call them “roof” and “ground” forces.

Here, then, are the two parameters for the roof firing upon the ground force.

Since pictures are quite powerful, let me show you the results up front:

I claim that the roof forces can hit the ground forces 50 m away, while the ground forces have to get within 30 m to return fire.

We have a formula for the angle of launch which leads to maximum range for the roof forces.

The angle for maximum range is noticeably less than 45°. Let me remind you what those other rules are that I used:

The first rule is the solution for the angle of launch which leads to maximum range. We worked that out in the previous post, using calculus. The second rule merely sets the acceleration of gravity.

Now we put that angle into our equations for sx and sy at maximum range… solve the second equation for the time(s) T when the projectile is at h = -10 m… and then get the horizontal distance for each of those times from the first equation…

That is, the projectile is at -10 m vertically at approximately –0.6 and +3.2 seconds, and the two distances from the roof are approximately –10 m and +50 m). Let’s look at that. I need the trajectory, i.e. the position vector as a function of time. Here it is with units:

Let me remind you what all those pieces were. fe2 was the general equations for the x- and y-components of position and velocity:

BC and BC2 are boundary conditions, to set the initial position to (0,0) and to express the initial velocity in terms of initial speed vo and angle of launch \alpha\ :

After applying the boundary conditions we have

The [[{1,2}]] in the following screenshot selects the first two equations – the curly brackets {} inside are crucial; it’s they that ask for elements of a list, in this case the first two.

Then s1, ps3, and cs, which we’ve seen already, set the angle, set the problem-specific parameters, and set the acceleration of gravity g.

Anyway, I want to plot the right hand sides of the final form of those two equations.

I have one last thing to do: get rid of the units (“strip” is a rule that sets Meter and Second to 1 – in fact, it also sets Feet and Foot to 1… in case I use English units, and because Mathematica® recognizes both Foot and Feet):

As a quick check, one would want the maximum altitude — and don’t call it h, because we’re using h already! Oh, I’ll remind you that we can get the maximum altitude by equating specific potential energy gh (i.e. mass m = 1) to the specific kinetic energy due to the initial vertical component of velocity. (We saw that in the first projectile post.)

So the maximum altitude is 8 m above the roof.

Rather than type in the values for times and distances, I really should let Mathematica compute them (called t3, t4, x3 for no particular reason):

The red dot marks the position of the negative-time solution. The point is that there is nothing special about time = 0 on the trajectory. Yes, we want it to be the time of launch from the roof – but this projectile could have been launched from the red dot at the negative-time solution. And no, it would not have the same velocity or angle of launch at the red dot as at the roof – but it could be launched to traverse the same path as if fired from the roof. If you couldn’t see the launch point, for example, you couldn’t tell whether the projectile came from the roof or from the red dot – or any other out-of-sight point on the trajectory.

I’m curious. Is that trajectory maximum range for a launch at the red point? To answer that, I need the velocity at launch. (Equivalently, I need the tangent vector of the trajectory at launch.) Here are the velocity equations… and I plug in \alpha\ , g, and vo – and change t to T… and then plug in the negative time solution…

Now I have equations in t – but a solution in T:

and it would be cleaner to just work with the RHS, changing t to T (otherwise sx[t] becomes
sx[–0.633682], which is OK but ugly):

Just the fact that the two components are different tells me that the tangent vector is not at 45°: this trajectory is not maximum range for a launch from the ground.

Let me get a clean picture of the trajectory. Note that I have changed the coordinate system simply by adding a constant vector {0,10} to the computed trajectory; I have thereby moved the launch point to (0,10) – my origin is now at the base of the building. Note also that I have explicitly set the image size and the plot range – and I changed the aspect ratio to correspond better with the chosen plot range.

That’s half the problem. Now set h = +10 m for the ground force firing upon the roof.

Our initial angle becomes:

The angle for maximum range is significantly more than 45°.

Now we put that angle into our equations for sx and sy at maximum range. As before, we solve the second equation for time T, getting two answers, and then we solve the first equation for the two corresponding distances.

That is, the projectile is at +10 m vertically at two times (approx 0.8 and 2.5 seconds) at two distances from the launch (approx 9 m and 29 m). The roof can hit the ground 50 m away; the ground force can return fire 29 m away. That’s not good: the ground force is vulnerable and unable to return fire for 21 m.

Let’s see that. I need the trajectory, i.e. the position as a function of time. Here it is with units:

so I want to plot the right hand sides.

Get rid of the units using “strip” to set Meter and Second to 1):

Just for the fun of it, again, get the maximum altitude:

Note that the maximum altitude isn’t all that much over 10 m.

Anyway, here’s what the ground force is doing:

As before, I use the red dot to mark the first time when the projectile is at the target altitude. In this case it’s part of the physical solution (“on the way up”) whereas in the other case it was at a negative time.

So. Would it be fair to say that the only reason why the ranges are different is because one is in the air longer than the other? Maybe. Almost. Essentially.

Maybe it would be too pedantic to say that that’s a very, very good first approximation; there is another, less significant factor: we change the angle of launch, too.

Let me close by putting the two trajectories on the same picture. In addition to using the same plot range and image size and aspect ratio, I have reflected and shifted the trajectory. The expression {-1,1} traj is a termwise product: it multiplies the x-component of traj by -1 and the y-component by +1. I also added a constant vector {x3,0} to the trajectory, so that the launch point is now (x3,0). In other words, my origin is at the base of the building, just as it was for the other revised plot.

Now I can put the two images together, and end as I began:

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