What is the optimal angle at which to throw an object?

I’m going to have to go back to my Physics book on this one…

45 degrees.

45 degrees applies if the ball is thrown from ground level. This ball is thrown from a distance of H above ground level, so I think it would vary.

Gimme a couple of minutes on this.

@wenn that’s what I thought I remembered from a year ago, but would that give you maximum distance. I think 45 degrees will yield maximum height and distance, but if you’re just going for maximum distance I’m not sure…I have to look this up. No one tell my Physics professor; he will not doubt think I should remember this, lol!

@yankeetooter 45 degrees yields maximum distance, 90 degrees yields maximum height. But I think the fact that the ball is not thrown from ground level changes things in this problem.

@PhiNotPi If this isn’t homework, then did you make up the problem? Do you definitely want A in terms of D, F, and H? Because the F feels problematic to me without knowing more info about your throw.

Yeah, @Mariah , as I’m googling this, I keep getting 45 degrees…I’m not sure the initial height of the throw would matter. The object will no doubt travel further than if it was thrown from ground level, but the question is about the angle needed from that height. So any results will all be taken from that initial height. Therefore, I believe that the answer would still be 45 degrees.

The object will no doubt travel further because of being thrown from a height above ground level, but ultimately 45 degrees should yield a further distance than any other throw from that height at a different angle.

@yankeetooter You’re probably right, but since it’s not coming intuitively to me for certain, I’m gonna try and do a proof. I could use the practice anyway.

Nothing like a good Physics question to get everyone’s brains working on a Sunday afternoon…

@Mariah I’ve been thinking about this problem for awhile now. I think I first thought about this problem when I built some sort of miniature-catapult-like machine. I knew about 45 degrees at ground level, but nothing is ever launched from ground level.

Theoretically it’s 45 degrees, but the actual answer depends on the wind.

@PhiNotPi That’s true. Even if a person is throwing the object from ground level, the object will launch from whatever height their arm launches it from. (Unless they’re standing in a hole, lol!) @jerv I don’t think we’re taking the wind into account here, according to @PhiNotPi ‘s question. We should assume a vacuum unless otherwise stated. I don’t think we want to have to fool with air resistance, or an inconsistently shaped object either…I know my brain doesn’t want to tackle this.

@yankeetooter I didn’t know if this question was a practical one or merely an academic one.

Ideally, Range = ( v^2 / g ) * sin (2 * angle) , where v = Initial Velocity and g = gravity (~9.8 m/sec^2 here on Earth)

@jerv I’m not sure it’s that simple. I remember having to take both vertical components (due to gravity) and horizontal components (due to the inital velocity of the object given by the thrower) into account, and then figuring the angle from there. It looks like you’ve only taken the force of gravity into account, which would only apply if the object was dropped. If the object was dropped, we wouldn’t be asking about angles…

Oh boy this is just beyond my Fluthergrade…

@yankeetooter What goes up must come down. The range is determined in part by the object’s horizontal velocity and in part by how long it takes for gravity to bring it back down. In order for gravity to pull it down, gravity first has to cancel it’s upwards velocity and then accelerate it downwards.

Assuming that the vertical component of the velocity is the same, range depends solely on the horizontal speed; the time it takes to make an object rise a certain distance and then fall back down depends solely on the vertical speed.

Since velocity is a vector quantity, the v term already includes the horizontal component.

Okay, so your throw of constant force accelerates the ball for as long as you’re in contact with it, but we don’t have that info. So I will simply call the velocity of the throw v0.
v0y is the y component of the velocity. v0x is the x component, which, ignoring air resistance, stays constant.
It hits its maximum height when vy = 0. So we can find the time at which that occurs using the equation, vy = v0y +a*t. vy = 0, a = -g, so v0y/g = t. This is to get to the top. To get back to height H, it takes double that time: 2v0y/g.
So it then has to fall a distance – H with the initial velocity -v0y. -H = -v0y*t + .5*-g*t^2. This requires the quadratic formula, (-b +/ – sqrt(b^2–4*a*c))/(2*a). t = (v0y – sqrt(v0y^2 + 2gH))/-g. Total time of trajectory is therefore 2v0y/g – (v0y – sqrt(v0y^2 + 2gH))/g = (v0y + sqrt(v0y^2 +2gH))/g.
We want to maximize distance in the x direction, which is equal to v0x*time.
D = (v0x*v0y + v0x*sqrt(v0y^2 + 2gH))/g = (v0^2*cos(A)*sin(A) + v0*cos(A)*sqrt(v0^2*sin(A)^2 + 2gH))/g
Now I will work on optimizing that shit. Lmao. This problem is harder to prove than I was expecting. Please correct me if you spot mistakes; this is so messy it’d be easy to make them. And I’d hate to find an answer and then see that I screwed up way back at the beginning.

If the problem was as simple as a lot of the earlier posters suggested, the military wouldn’t have spent loads of money in the past to create firing solution calculators to work this out. If you are after maximum distance, you also have to take into account the Coriolis forces acting on the projectile at the point of launch causes by the rotation on the Earth.

I’ll have to look at your response in more detail later when I’m not falling asleep, but it looks like what we used to do in Physics class last year, where we took both horizontal and vertical components into account.

@Tocon_Tactus Too complicated for me at this point, lol!

@Tocon_Tactus Things like air resistance, Coriolis forces, etcetera are the difference between “textbook” and Applied physics.

Yes, we never took it this far in class…thank goodness.

So if the distance that I found is right, its derivative is cos(A)^2 – sin(A)^2 + (sin(A)cos(A))/sqrt(sin(A)^2 +2)). WolframAlpha shows a maximum point at about 1 radian, which is about 37 degrees. I think I probably screwed up somewhere.

This is a much harder problem to prove than it appeared to be! I think I’ll probably come back and try it again later with a fresh brain.

Oh yeah, and I replaced all the constants with one to simplify doing the derivative, but now I’m realizing that that wasn’t okay to do. Especially because H wasn’t a constant and I replaced it with one also, whoooooops. My above response is most definitely incorrect.

What is the the force (speed) constant? I think that in order to calculate that optimal angle, you need to know the exact force/speed at which the ball will be traveling.

If we know the vertical velocity, we can find the height where the vertical velocity is zero. We can then find the time that it takes to fall from the maximun height to the launch height, and add this to the time that it will take to fall from the max height to the ground, we will get the time that it takes for the object to hit the ground from the time of launch. Given this time, we can find the distance that the ball goes horizontally by just multiplying the time with the horizontal velocity.

@everyone Yes, I should have put constant velocity instead of constant force in the question.

@PhiNotPi I don’t think that your method is valid. One has to do it using vectors, and can’t just multiply the time with the horizontal velocity. Remember that the vertical force, gravity is an acceleration, whereas the horizontal velocity would, under ideal conditions, be constant.

Unless you’re throwing from a hole in the ground, which means that you may need to throw with a vertical angle sufficiently high to clear the wall of the hole you’re in, then 45° is always the optimum “textbook” angle, as @jerv correctly says. If you’re having to deal with a strong headwind, then a lower angle will be better. Contrariwise, if you’re dealing with a strong tail wind, then a higher angle would work better to keep the object aloft longer.

@yankeetooter No, you can multiply horizontal velocity by time. Because horizontal velocity is constant, this is valid. But in order to find the time, you have to use information from the vertical component of the flight.

It’s sounding like 45 degrees is the optimal angle whether you’re above ground or not, so I apologize for expressing doubt earlier in the thread. It’s just driving me nuts that I couldn’t prove it.

If you ignore wind resistance and ballistics it is 45 deg. However if you want to push the envelope….
The answer is much more complicated depending upon the speed, density and drag coefficient of the object. For a supersonic bullet, best angle for furthest distance is typically between 35 and 40 degrees. There are many trajectory simulators on line you can try.
Stick with 45 deg if you are throwing it.

You guys, I really hate to be obnoxious about this but unless I am making major math mistakes, it definitely isn’t 45 degrees.
To simplify the work, I substituted in arbitrary values for initial velocity and for H. I set v0 = 2 and H = 3. I also substituted 9.8 for g. If 45 degrees were always the optimal angle, it wouldn’t matter what values I give these parameters; 45 would always yield the farthest distance.

Again, first I’ll find the time it takes to reach the top of the trajectory, where the y component of the velocity = 0.
v = v0 + a*t
0 = 2sin(A) – 9.8*t
(2*sin(A))/9.8 = t
It takes double the time to return to a height of H, so (4*sin(A))/9.8.
Now I’ll solve for the time it takes to fall to the ground from height H (I set H = 3)
d = v0*t +.5*a*t^2
-3 = -2sin(A)* t + .5*-9.8*t^2
Using the quadratic formula, obtain t = (2sin(A) – sqrt(4sin(A)^2 + 58.8))/(-9.8)
So total time of flight is (4*sin(A))/9.8 + (2sin(A) – sqrt(4sin(A)^2 + 58.8))/(-9.8) which simplifies to (2sin(A) + sqrt(4sin(A)^2 + 58.8))/9.8
Velocity in the x direction is constant, so Vx = D/t
2cos(A) * (2sin(A) + sqrt(4sin(A)^2 + 58.8))/9.8 = D, or
(4cos(A)sin(A) + 2cos(A)sqrt(4sin(A)^2 + 58.8))/9.8 = D
So when I plug in 45 for A, I should get the highest possible value for D. BUT, when I plug in 45, I get approximately 1.329. When I plug in 44, I get about 1.348.

So have I screwed up? Or is there more to this problem than we’re assuming?

@Mariah I am not sure if this could be the problem, but when you calculate the time it takes to fall from height H, are you taking into account the fact that the object would already have a downward velocity at height H?

To possibly simplify the problem, instead of calculating the time it takes to fall back to H from the top, and then calculate the time it takes to fall from H to the ground, just calculate the time it takes to fall from the top to the ground.

@PhiNotPi Yeah, its velocity there would be -v0y, that is, it’s falling downwards at the same rate you launched it.

I’ll try doing it again with the method you mentioned and see if I get the same result.

Yup, same result, considerably less complicated math. Thanks for the suggestion – it made things a lot easier and now I feel fairly confident that I haven’t made a mistake.
So as I found earlier, the time up to the top of the trajectory is (2*sin(A))/9.8.
Now we have to determine how high up the top of the trajectory is.
d = v0*t + .5*a*t^2
d = 2*sin(A) * ((2*sin(A))/9.8) – .5*9.8*((2*sin(A))/9.8)^2 which simplifies to (2*sin(A)^2)/9.8
So the total distance it has to fall is (2*sin(A)^2)/9.8 + 3 (I set H = 3).
v0 = 0 here because the vertical component of the velocity is 0 at the top of the trajectory.
d = v0*t + .5*a*t^2
(2*sin(A)^2)/9.8 + 3 = 0*t + .5*9.8*t^2
t = sqrt((2*sin(A)^2/48.02) + 3/(4.9))
So the total time of flight is (2*sin(A))/9.8 (time to top of path) + sqrt((2*sin(A)^2/48.02) + 3/(4.9)) (time all the way back down).
Again, total distance is just the constant horizontal velocity (2cos(A)) times total time.
D = (4*sin(A)*cos(A))/9.8 + 2cos(A)*sqrt((2*sin(A)^2/48.02) + 3/(4.9))
When I plug in 45 I get approx 1.329, 44 yields 1.348. Same as before!

I would really like for someone who says it’s always 45 degrees to cite an article or something, because my math certainly isn’t infalliable but I’ve gotten the same differing answer twice using two different methods now, so I feel pretty convinced.

Here’s a page that will do the math and allow you to compare results. Go to “Where Will It Land?” near the bottom of the page.

That’s pretty awesome @WasCy !

@WasCy Wow, that’s awesome! And the page’s results were the same as mine for v0 = 2, H = 3, and A = 44 or 45.

However, we have yet to answer @PhiNotPi‘s original question. If it’s not 45 degrees, what is it? I’ll work on it again tomorrow! Or later tonight if it keeps me awake. This kind of thing tends to do that to me.

I wish I could take credit for the awesomeness of the page. The only awesomeness I can claim is my own, and I have to share that with so many others… those giants whose shoulders I have stood upon, and the other giants who’ve carried me on their backs for so long. Not to mention the giants who birthed me.

Clever…and humble! Lol!

Get the equation for range vs angle. Should be a parabolic equation. Max is when slope = 0.

@RocketGuy The trouble is, that equation is in terms of v0, A, and H, and I have yet to study multivariable calculus so I can’t differentiate to find where the slope is zero. Anyone up for optimizing this?

Range = (v0^2*cos(A)*sin(A) + v0*cos(A)*sqrt((v0*sin(A))^2 + 2gH))/g

How about making a graph on Excel?

Wouldn’t it have to be a four-dimensional graph? :\

@RocketGuy : graphs on Excel, ugh! I had enough of this last semester when doing lab reports. At least I learned how to do it…

Graph: Range vs Angle A. Keep v0 and H constant.

@RocketGuy Optimal angle definitely will vary depending on v0 and H, but maybe if I did several Range vs. Angle graphs with several values of v0 and H, a pattern will emerge. Hm.

I think it will vary with H, but not with v0. I did this calculation many years ago, with H=0 => v0 drops out.

I just thought of a way to find the optimal angle. Given the range, you can calculate two different launch angles. The optimal range would be the one that causes these two angle solutions to converge and become equal. Once we know the optimal range, we can find the optimal angle that will give this range.

So, here is how we could go about solving it. First, we need the equation for Range in terms of Velocity, Angle, and Height. Then, we need to convert this to an equation for Angle in terms of Velocity, Range and Height. Somewhere inside this new equation there will be a +/- sign, which is what causes the two solutions. To make the two angles converge, the stuff inside of the +/- sign needs to equal 0. This “stuff” will be an expression involving H, V, and/or R. H and V are fixed, so we need to solve for the R that will make this expression equal zero. This has to be the optimal R. Once we find the optimal R, we can just plug it back into the larger equation to find the optimal angle.

Turning this into a straight-foward equation for optimal A might be hard. It will involve writing optimal R in terms of V and H (using the stuff in the +/-) and then substituting this into the equation for A in terms of V, H, and R. This will give us the the equation for optimal A in terms of V and H.

I hope that this approach will give us what we are looking for.

@RocketGuy Right, when H=0, R = (2*v0^2*sin(A)*cos(A))/g which is very easy to analyze – R always increases as v0 increases, and R is optimized when sin(A)*cos(A) is at a maximum – this occurs when A = 45, regardless of v0. By the way, I feel the need to add a disclaimer because I feel so pretentious right now – I know you don’t need it explained, I just have a tendency to think out loud, haha. The equation for R when H > 0 is much less intuitive and I don’t feel 100% certain that I can assume that v0 doesn’t impact the optimal angle this time. I’m not sure. I’m going to do some experimenting on the website that @WasCy linked, input random values for v0 and H and then test a bunch of angles to see what whole number is optimal. Then I’ll wildly change v0 and see if optimal A is still the same. That’d be enough to convince me even though it’s not a good formal proof.

@PhiNotPi Very interesting – I think I’m going to have to read it through two or three more times to fully understand, and then I’ll look into doing it. Finding this as a function of A will indeed be very difficult.

I just wanted to share some more thoughts on this that I concluded after staring at it for a while last night.

Range = (v0^2*cos(A)*sin(A) + v0*cos(A)*sqrt((v0*sin(A))^2 + 2gH))/g

Note that if you set H = 0 the equation reduces to R= (2*v0^2*sin(A)*cos(A))/g where optimal angle can be determined by simply figuring out what angle optimizes sin(A)*cos(A). The above formula is much less intuitive. The last term, which includes a sine, has some constant added to it before it is combined with other terms. After thinking about it for a while, I concluded that this makes sine the less “important” trig function for optimizing R. My reasoning is this. Multiplying two “medium” numbers together gets you a higher result than multiplying a small and a large number (ie. 7*7 > 6*8), so if you have to add something to one of your factors, it is more beneficial to add to the smaller one: you get a higher product if you raise your smaller factor than if you further raise your larger one. So, since you’re adding a term onto one of your sine terms, sin(A) needs to be smaller than cos(A), causing the optimal angle to be below 45 degrees (because angles below 45 have higher cosines than sines). I hope my reasoning is sound and I explained it that okay. So, as H increases, the optimal angle decreases, because the difference between sine and cosine becomes more profound as the term added to sine becomes larger.

Gonna go do some more thinking, now – I’ll definitely look into your ideas @RocketGuy and @PhiNotPi. Thanks!

Dang it – v0 definitely impacts optimal angle when H > 0. I went back to @WasCy‘s link and input v0 = 2 and H = 3. The optimal (whole number) angle for that case was 14 degrees. Then I set v0 = 10, leaving H = 3, and the optimal whole number angle jumped up to 38 degrees. Too bad – I was hoping we’d be able to eliminate one variable.

@Mariah It makes sense (not much real sense, but mathematical sense) that v0 would affect the optimal angle. To prove this, I need to use the scaling principle. When you increase the height, you are in fact scaling up the problem. If you double the height, every length should be multiplied by two, and the math will remain the same. The thing is: Velocity has a length component. Velocity is directionalized length/time. So, when you multiply the height by a number, you have to multiply the velocity by that same number. If you don’t, the optimal angle will change, because you are no longer scaling the problem. Doubling the velocity is the same as halving the height, when it comes to finding the optimal angle.