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:
$\cos2\theta=\frac{gh}{v^2+gh}$
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.
Best Answer
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.
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.
So the best angle needs to depend on velocity as well as the height of projection.