A while back, I was curious about how the optimal angle (ie. max range) for a projectile (with no drag and so forth, so HS physics conditions) would change if it is thrown from a height $h$.

So I started working on this, and about $1.5$ pages into my $4$ page solution [full of ugly trig and differentiation, I did not think to use Wolfram :( ...], I decided to assume $v$ (the velocity with which projectile is thrown) to be equal to $\sqrt{g}$. I did this because this simplified the trig a bit and I thought that the optimal angle would be independent of $v$. Because it is well known that when $h=0$, optimal angle is $\theta=45^\circ$ irrespective of $v$, so I thought the same would hold for any arbitratry $h$. So after some more messy math I got optimal angle $\theta$ is:

$\sin^2\theta = \frac{1}{2(h+1)} \iff \cos2\theta = \frac{h}{h+1}$

Then I looked up the solution and found this Physics Stack Exchange answer, which gave the answer:


So my answer is the correct when $v=\sqrt g$ but I was surprised when I saw that the answer depends on launch velocity. Of course I looked at the math again (with my good friend Wolfram, this time) to see that the answer did infact depend on $v$ also. But I still do not understand why this is the case. I mean this is not the case when $h=0$. Also, it seems to me like increasing the velocity would just scale the answer by a positive factor, so I am not sure why the velocity is important here.

So, Tldr; I want to understand intuitively/qualitatively (I understand it mathematically) why the optimal angle for maximising range in a projectile is not independent of the launch velocity.

  • 6
    $\begingroup$ Note that the expressions $v=\sqrt{g}$ and (in your denominator) $h+1$ are both dimensionally inconsistent: you've made at least one algebra error in your path to those impossible statements. In the expression you quote, the product $gh$ does have the same units as the square of a speed. $\endgroup$
    – rob
    Commented Oct 5, 2021 at 17:02
  • $\begingroup$ @rob , I meant $v=\sqrt g$ in magnitude, so $v = \sqrt (9.8) \frac{m}{s}$ $\endgroup$ Commented Oct 5, 2021 at 17:04
  • 9
    $\begingroup$ But $v$ and $g$ don’t have meaningful magnitudes unless you specify a system of units. When I say something like “the distance from my house to downtown is twenty minutes,” that only makes sense if I also specify some distance-to-time conversion factor (here, typical highway driving speed). If you have written $v=\sqrt g$ to mean $v = \sqrt{\left(9.8\frac{\rm m^2}{\rm s^2}\right)}$, you have implicitly multiplied by $h=1\rm\,m$ somewhere. The most common way this happens is that you accidentally write $g$ instead of $gh$ and don’t notice your error. $\endgroup$
    – rob
    Commented Oct 5, 2021 at 19:04
  • $\begingroup$ @rob I think we are having a misunderstanding ;). What I mean is at some point, I got the equation $h = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}$ where $x$ is in meters. For now, this equation doesn't really have much value apart from the fact that it is clearly dimensionally consistent. So now, to cut down on the algebra, I let $v=|\sqrt g| \frac ms$. So then the equation was reduced to $h= x tan\theta - \frac{x^2}{2cos^2\theta}$. This is the same equation but I just did not bother writing out the units. Much like how for uniform acceleration, $s_n = v_0 + \frac12a(2n-1)$ is dist. contd... $\endgroup$ Commented Oct 6, 2021 at 16:21
  • 3
    $\begingroup$ That's the error: if $x$ is dimensionful, the difference $x\tan\theta - x^2/\cos^2\theta$ is not allowed. $\endgroup$
    – rob
    Commented Oct 6, 2021 at 16:43

5 Answers 5


Let's take an extreme situation where $h$ is very high.

According to both formula, the best angle should be zero, especially if $v$ is low, so path (1) gives the best range.

If the angle were independent of velocity it would be zero degrees even if $v$ were increased.

enter image description here

But if $v$ becomes very large, so that $h$ looks small, it's better for the angle to be like path (2). The projectile is now spending nearly all the time above $h$.

This is similar to a projectile launched from a lower height - an angle near 45 degrees would be best.

enter image description here

So the best angle needs to depend on velocity as well as the height of projection.


The basic reason this happens is that by adding in a length scale $h$ into the problem you have increased the number of dimensionful parameters the answer can depend upon. When $h=0$ the maximum range occurs at a (dimensionless) angle that can only depend upon

$$ \theta_{\rm Max}=\theta(v,g) \ , $$ but there are no dimensionless combinations of $v,g$ that it could possibly be. Thus it must be independent of both.

However now there is a length scale $\theta_{\rm Max}$ depends upon

$$ \theta_{\rm Max}=\theta(v,g,h) \ , $$

and must depend upon these only via the dimensionless ratio

$$ \Pi = \frac{v^2}{2gh} .$$

For completeness - we can work out the limits of what the function $\theta$ must be - at very large $\Pi$ we have a very fast projectile from a negligibly small tower - the maximum range must be obtained at $\theta=45^\circ$. At $\Pi \ll 1$ we instead have a very tall tower such that adding a vertical component to the velocity won't really affect the fall time, so we should just fire horizontally. Thus the function

$$\theta_{\rm Max}(\Pi)$$ that we are after is $0$ for $\Pi=0$ and $45^\circ$ for $\Pi\gg 1$. The function $$\theta_{\rm Max}(\Pi)=\frac{1}{2}\arccos \frac{1}{2\Pi+1}$$ from the OP has exactly these properties.


Adding the following fact that I find curious.

Assume that air drag can be ignored. Assume that the launch speed and height are known. Then the launch angle yielding the maximum range is such that the initial speed vector and the landing speed vector are perpendicular to each other.

The proof is relatively simple. Assume that the initial speed is $\vec{v}_{init}=(v_1,v_2)$. High school physics then tells us that

  • the duration of the flight is $$T=\frac1g(v_2+\sqrt{2gh+v_2^2}),$$
  • the landing speed is $\vec{v}_{final}=(v_1,-\sqrt{2gh+v_2^2})$, and
  • the range is given by the formula $$s=Tv_2=\frac1g\left(v_1v_2+v_1\sqrt{2gh+v_2^2}\right).$$

Let us introduce the somewhat unphysical (may be just unusual?) quantity, the inner product of the initial and the final speeds: $$ D:=\vec{v}_{init}\cdot\vec{v}_{final}=v_1^2-v_2\sqrt{2gh+v_2^2}.$$ Straight forward (junior high school) algebra then shows that $$ \begin{aligned} (gs)^2+D^2&=(v_1^4+2v_1^2v_2^2+v_2^4)+2gh(v_1^2+v_2^2)\\ &=||\vec{v}_{init}||^4+2gh ||\vec{v}_{init}||^2. \end{aligned} $$ The claim follows from this. Maximizing $s$ is equivalent to maximizing $gs$. On the right hand side of the last equation everything is a constant determined by initial constraints. Most importantly, it does not depend on the launch angle. Therefore maximizing $gs$ is equivalent to minimizing $D^2$. In other words, we require $D=0$. This, in turn, is equivalent to $\vec{v}_{init}\perp\vec{v}_{final}$. QED.

  • Launching the projectile near vertically results in a negative value for $D$. Similarly, launching near horizontally leads to $D>0$. By continuity, there is a launch angle such that $D=0$ is achieved.
  • Assume $h>0$. If the launch angle is 45 degrees or higher, the projectile will be descending at that same angle at the point, where it has descended back to the initial height $h$. Gravity will make any further descent steeper, resulting in an angle exceeding 90 degrees between $\vec{v}_{init}$ and $\vec{v}_{final}$. Therefore the optimal launch angle is below 45 degrees. Similarly, if $h<0$ we see that the optimal launch angle is above 45 degrees.
  • This also answers the title question. Launching at the same angle with a higher velocity amounts to multiplying both $v_1$ and $v_2$ by the same constant $>1$. From the formula for $D$ you see that $D$ won't be multiplied by the same constant. The presence of that $2gh$ under the square root means that the negative term is affected less. Resulting in $D>0$.
  • We also arrive at the following formula for the maximum range: $$s_{max}=\frac1g||\vec{v}_{init}||\sqrt{2gh+||\vec{v}_{init}||^2}.$$
  • $\begingroup$ The claim was recently brought up in a Finnish Facebook group. The above is my proof for it, streamlined as much as I could manage. I am inexperienced at Physics.SE, so I apologize, if I committed a faux pas. $\endgroup$ Commented Oct 31, 2021 at 7:21
  • $\begingroup$ This is a good answer. Welcome! $\endgroup$ Commented Oct 31, 2021 at 7:44
  • $\begingroup$ Thanks for the nod @EmilioPisanty! Most appreciated. I'm still somewhat concerned. My exposure to math tricks exceeds my familiarity with physics problems by at least three orders of magnitude. So I would not have been at all surprised to find out that seasoned physicists have covered this little curiosity many times over on this site. My search came out empty, so I chose to post it. Possibly because I'm a bit too proud about having figured it out :-/ $\endgroup$ Commented Oct 31, 2021 at 9:22

Consider that in projectile motion, the initial angle determines the shape of the parabolic path followed by the projectile and the initial velocity determines the size. A lager velocity gives a greater maximum range, but not the shape of the path. (Changing the velocity is like changing the numbers along the x and y axis.)


Continue the projectile motion backwards in time as if it were launched from the ground at an earlier instant. This gives a virtual launch point using a virtual launch angle and virtual velocity. You now have parabolic motion from the virtual launch, through the real launch, then back to the ground.

We know that the virtual launch velocity and virtual launch angle must produce projectile motion having the real launch velocity and real launch angle when the projectile reaches the real launch point.

By inventing this virtual launch, we are able to bring in our usual understanding of parabolic projectile motion.

If we fix the real launch angle, different choices of the real launch velocity pick different virtual launch points. Higher velocities move the virtual launch point further away from the real launch point and lower velocities move it closer. The same is happening to the terminal point, so the total horizontal displacement depends on the real launch velocity.

If we fix the real launch velocity, lower real launch angles move the virtual launch point further away and higher real angles move the virtual launch point closer. This isn't symmetric -- compare what happens when the real launch angle is nearly vertical (lowering the angle moves virtual launch and terminal point away) with what happens when the real launch angle is nearly horizontal (increasing the angle moves the virtual launch point closer and the terminal point further away).

These choices correspond to placing the real launch point at various places on the virtual parabolic trajectory. To maximize the horizontal displacement for the full virtual trajectory, the virtual launch angle is $45^\circ$. However, as we move along that trajectory, the angle and velocity are functions of horizontal displacement. We can eliminate the displacement between them and get that (for that point on the trajectory) the real launch angle depends on the real launch velocity (and vice versa).

  • $\begingroup$ This is a great answer, Thanks! $\endgroup$ Commented Oct 7, 2021 at 19:33
  • $\begingroup$ "If we fix the real launch velocity, lower real launch angles move the virtual launch point further away and higher real angles move the virtual launch point closer. Again, by symmetry, the same is happening to the terminal point." There is no symmetry here. If you start with horizontal launch and then slightly raise the angle, the virtual launch point will move closer, but the terminal point will move away. As a result, the max distance from the real launch point and the terminal point happens at a different launch angle than the max distance from the vir.tlaunch point and the terminal point. $\endgroup$
    – Litho
    Commented Oct 8, 2021 at 12:08
  • $\begingroup$ @Litho : Yup. Got a little turned around there. Fixing. $\endgroup$ Commented Oct 8, 2021 at 21:01
  • $\begingroup$ Without the symmetry, your last paragraph doesn't work. We're not looking for the trajectory which maximizes the distance between the virt. launch point and the term. point; we're looking for the trajectory which maximizes the distance between the real launch point and the term. point. The two trajectories would be the same if there was symmetry, i.e., if the term. point moved away from the real launch point whenever the virt. launch point moved away, and vice versa. But it is not so, and the two trajectories are different. (cont) $\endgroup$
    – Litho
    Commented Oct 9, 2021 at 7:44
  • $\begingroup$ So we cannot just move the real launch point along some optimal virtual trajectory. If a virtual trajectory has the optimal angle for the height and speed at one launch point on it, it does not have the optimal angle for the height and speed at its other points. $\endgroup$
    – Litho
    Commented Oct 9, 2021 at 7:48

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