Physical explanation for different angles yielding the same range in projectile motion It's well known that throwing a ball with a given speed at angles alpha or (90-alpha) will get it to land in the same distance. It's easy to see from the equations, but is there a more physical explanation for this? 
 A: The simple explanation: if you throw a ball at a low angle, it travels quickly in the horizontal direction, but for a short time. If you throw it more vertically, it has little horizontal velocity but spends much longer in the air.
Either of those two can get you to the same distance. Mathematically, if initial velocity is $v$ at angle $\theta$ to the horizontal, you have
$$\begin{align}v_h &= v\cos\theta\\
v_v&=v\sin\theta\\
t &= \frac{2 v_v}{g}\\
d &= v_h t \\&= \frac{2v\sin\theta\; v\cos\theta }{g}\\
&=\frac{v^2 \sin 2\theta}{g}\end{align}$$
Of course $\sin 2\theta$ peaks at 45° ($\pi/4$), and has the same value at $\pi/4 + \phi$ and $\pi/4 - \phi$ for any $\phi$.  You can see that both intuitively, and mathematically, there will be two solutions that give the same distance (except in the case that $\theta=\pi/4$ where the two solutions are the same).
A: Yaron has asked for a "physical" explanation, but there is a "mathematical" one that is important enough that every physicist should be familiar with. I put scare quotes around those words, because the distinction is not as clear as some people would like it to be. In this case the core "mathematical" fact is predicated on a physical assumption about the nature of motion.
Let's start by establishing the behavior at a couple of important special case, and in both cases I shall assume that we throw the ball from zero height, and I shall define the range as the horizontal displacement from the point of release of the first point after release at which the ball has height zero.1


*

*Horizontal throw: ($\theta = 0^\circ$) The ball starts at height zero and is released moving horizontally and so it immediately at height zero. Range is 0.

*Vertical throw: ($\theta = 90^\circ$) The ball goes straight up and falls straight back down landing at the point from which it was launched. Range is 0.

*Maximum range ($\theta = \theta_{max}$) We accept from experience that there is some angle at which we get the maximum range.
Now we add an important physical fact: the motion of the ball is (mathematically) continuous. From this we conclude that the range as a function of angle must be a continuous function.2
At that point we can pick any range achieved at low angle ($0^\circ < \theta < \theta_{max}$) and the intermediate value theorem tells us there must be a high angle ( $\theta_{max} < \theta < 90^\circ$) with the same range, and this argument can be turned around so show that for angle high angle we can find a low angle with matching range.
The only physical fact we rely upon here are that position (and velocity if there are forces depending on it) is a continuous function of time.

These kinds of arguments (relying on the continuity of various physical quantities under real conditions) are useful for establishing expectations for many simple systems.

1 I'm also assuming no wind (but not necessarily not atmosphere3) and no inertial-pseudo forces that we need care about.
2 I think that https://math.stackexchange.com/a/430341/8422 covers the case of integrating the equations of motion.
3 The closed form analysis shown in some of the other answers is dependent on a zero air resistance condition, but this argument works fine with air resistance. It will even work in a constant breeze, but finding the zero-range limits is then non-trivial.
A: As you already said in your question it is indeed easy to see from the equation that the ball will travel the same distance when thrown under an angle $\theta $ or $(\pi/2 - \theta)$. But for the sake of completeness, and for clarity I shall go over it step by step. 
I'll start by defining the angle $\theta$ as the angle between the ground and the direction in which we throw. It will be usefull to realise that if $\theta$ is small $\pi/2-\theta $  will be big or vice versa. 
The problem can be devided into two parts. On the one side there is the vertical motion that can be described as a constant linear accelerated motion (with acceleration $a=-g$). On the other side there is the motion at constant speed in the horizontal direction. 
Obviously the ball, once thrown, will travel through the air until it hits the ground again. So the vertical motion determines how long the ball will travel through the air. Obviously this depends on the vertical speed of the ball (and thus on the angle under which the ball is thrown), as can be seen from the following equation: $$t = 2gv\sin(\theta)$$
This equation follows easily from the formula for speed in a constant linear accelerated motion. The physical meaning of this formula comes down to this: the higher the vertical component of the velocity of the ball the longer it will take for the ball to return down. This is also what we expect from day to day experience. 
Whilst the ball is clear of the ground (and moving through the air) it travels a certain horizontal distance which depends of the horizontal component of the speed: $$\Delta x = v\cos(\theta)t$$
Once again we can understand this formula by looking at our experience from throwing a ball of even driving the car. The higher our speed the further we can travel in a fixed time interval t. Now separated these two equations do not show the $\theta \leftrightarrow (\pi/2-\theta)$ symmetry. However when we combine the horizontal and the verical motion to derive the the actual travelled horizontal distance we get: $$\Delta x= \frac{2v^2}{g}\sin(\theta)\cos(\theta)$$
This equation remains the same under $\theta \leftrightarrow (\pi/2-\theta)$ exchange. This is because the sine tranforms in a cosine and vice versa. From the above discussion (with constant $v$!) we can see that the symmetry comes from a trade of between spending more time in the air (bigger vertical velocity component thus bigger $\sin(\theta)$ and thus a bigger $\theta$) but traveling slower in a horizontal direction, or spending less time in the air but moving faster horizontally (bigger vertical speed thus bigger $\cos(\theta)$ thus smaller $\theta$). 
A: There's lots of mathematical answers, I thought I'd post a purely non-math explanation.

If you throw the ball at a larger angle, it will stay in the air longer. If you throw the ball at a lesser angle, it will go to the right faster. There is a trade-off between time in the air and speed at which you are going to the right, so two balls thrown at different angles can end up at the same final location.
A: Absent air resistance, the maximum distance is achieved when $theta$ is 45.  So when the angle is more or less than 45, the distance is less then the maximum.  I don't think its obvious that the distance will vary as you describe, but note that when the ball goes straight up, or exactly horizontally, its goes the same distance (zero).
A: Think of baseball.
A long high fly to the shortstop will put you out just as surely as a line drive to shortstop.
