# Maximum range of a projectile launched from elevation “dumbed down”

I am trying to conceptually understand why the angle which produces the greatest range for a projectile launched with an elevation is not 45 degrees. I have exausted all other options, and I hope that someone here can explain it to me conceptually. That is, I am currently not at the mathematical level to explain this feature of projectiles using derivatives- but I am required to understand why this is the case.

The real source of my confusion is this app

The projectile in this application is launched from an elevation, and yet 45 degrees still produces the maximum range for the projectile! From my understanding the maximum range of projectile depends on configuring the launch to maximum amount of "hang-time." Is this application flawed - or am I going crazy.

As has already been mentioned in the other answers, a useful way to think about this problem is to consider the balance between the "hang time" you get from a vertical component of the velocity and the horizontal component that actually gets you some distance covered in however much "hang time" you have. What I'd like to add is a simple but powerful trick using very little mathematics* that I think may be helpful.

This trick is to consider a particular limiting case. If you imagine throwing a (soft, so nobody on the ground gets hurt) projectile from a very tall building, it is a pretty straightforward matter to convince yourself of the following: The additional "hang time" that you get from an upward component of the velocity will be small compared to the amount of time that the projectile will take to fall down the height of the tall building. This means that in order to cover the most ground, what you should really do is put all of the available energy into covering as much distance as possible, since you already have a large amount of "hang time" available to begin with. That is, the relative value of adding projectile time decreases.

Anyway, I hope this is a helpful way to think about it!

• I said very little mathematics, actually I did a bit of a trick, one that is useful in more circumstances than this. What I've done is called asymptotic analysis (examining the limiting behavior of the system). Often this can simplify the math quite a bit, and since the math wasn't too complicated to begin with, it's almost gone here!
• That is a awesome explanation! The concept of asymptotic analysis is certainly something I will take into consideration when attempting to understand concepts involving a limiting case. – Kurt Apr 20 '12 at 6:13
• Great! Glad you liked it. It's a very powerful tool, as it's sort of a way to "step back" from the problem and look at the big picture. It turns out to be useful for developing intuition about stuff like this, and can also be used to learn something about problems that are very difficult to solve directly. – tmac Apr 20 '12 at 6:23

Without mathematics, it gets hard. You don't really need derivatives; basic SUVAT equations are fine and you need to know some basic trig.

Anyway, I'll try:

It's not just maximum amount of "hang time"--you have two things fighting against each other here.

If the projectile has the same initial velocity, you can break it into two parts--horizontal and vertical.

The vertical velocity determines the "hang time"--for a fixed vertical component and elevation, the amount of time it stays in the air is fixed ($\frac{u_y +\sqrt{u_y^2+2gh}}{g}$ to be precise).

On the other hand, the horizontal velocity determines how fast it goes along the ground.

So, the more hang time (ie more vertical velocity), the longer we have to cover whatever range we want. But, the more horizontal velocity, the faster we can cover the range in a given hang time.

The issue is, there is a balance between the two-- they need to vector-add to the total velocity ($u_{tot}^2=u_x^2+u_y^2$), so we can't make them both arbitrarily large. Making one larger makes the other one smaller. And since the range depends directly upon the both of them, we need to find a balance.

When there is no elevation ($h=0$), the answer comes out to be exactly $45^o$--we get a $\sin2\theta$ term in the expression for range. I can't think of an intuitive explanation for this, I guess that the "balance" thing is enough.

At other elevations, the balance is different, since the hang time increases but the horizontal speed doesn't. So the balance "shifts".

Mathematically, the formula for range at elevation $h$(initial velocity $u$ at angle $\theta$) would be:

$$R=t_{hang}\times u_x=\frac{u_v +\sqrt{u_v^2+2gh}}{g}\times u_x=\frac{u^2 \cos\theta \left(\sin\theta +\sqrt{\sin^2 \theta +\frac{2gh}{u}}\right)}{g}$$

At h=0, this becomes the relatively simple $\frac{u^2\sin2\theta}{g}$--one can see that this will be maximum at $\theta=45^o$

On the other hand, the more complicated, general formula looks like the angle at which the expression is maximum will depend upon $h$--to be precise, on the value $\frac{2gh}{u}$--yep, it depends upon the initial speed as well.

Wolfram|Alpha's not listening to me, and I don't want to differentiate it, but I hope you get the point :) Anyway, you didn't want any mathematical part..

• Thanks a ton - that does help me to understand, and the math is comprehensible to me. That is, it looks as though Range at an elevation is calculated by finding the product of the x-compoenent of the initial velocity and "hang-time," which is calculated by using the quadratic equation. Thanks again- I am going to play with this until I have a better grasp – Kurt Apr 20 '12 at 5:39
• @Kurt: Yep. The main part is the presence of the $\frac{2gh}{u}$ deep inside the equation, which means that the value of $\theta_{max}$ is most probably dependent upon that expression. If it was outside the equation (added/multiplied to it), then it would be fine. – Manishearth Apr 20 '12 at 5:46

You're right that we want to maximize the hang time, but there is an additional constraint; we want to maximize the hang time to ensure we cover the most ground in the horizontal direction. Clearly, if we just want to maximize hang time, we just shoot straight up... but then we don't cover any ground. So, the problem at hand is, how do we cover the most ground (this stores in it the presupposition that: time has something to do with it)?

neglecting any acceleration in the horizontal direction, the distance the projectile covers will be dependent on how long it can travel with its given horizontal velocity. So we want to maximize the horizontal velocity, and hang time. Now notice that once we fire the projectile past an angle of $45^{\circ}$ from the horizontal, we start to sacrifice horizontal velocity for hang time. In the other way, if we fire under $45^{\circ}$ from the horizontal, we lose hang time in favor of a faster horizontal velocity. Is there any way to win?

Yes, there is an angle that gives us the most horizontal speed, for the most hang time, namely, $45^{\circ}$.

I hope this helps.

• Thanks, that does help cement by understanding of why a 45 degree angle maximizes range. But the real concept I am wrestling with is why, at elevations, 45 degrees no longer results in a maximum range. – Kurt Apr 20 '12 at 5:47

To conceptually understand this, think about the extreme case. Suppose you wanted to maximize the horizontal distance by launching a projectile over a canyon of infinite depth. What would be the angle you'd choose? Right, a horizontal launch would give you the most horizontal distance. So for a finite depth, the launch angle will be less than 45 degrees.

As a sideline: if the projectile lands at the same level it's launched, the launch angle and the impact angle are both 45 degrees ... adding up to a total of 90 degrees. The same is true for the horizontal launch over the infinite canyon ... in the limit the projectile will be falling straight down, or 90 degrees with the launch angle of zero degrees. Do you suppose that is true for any height ... that the launch angle and the landing angle add up to 90 degrees. Well, that's a math problem that might be interesting to solve with a little calculus.