Why is the range of a projectile proportional to the square of the initial velocity? I was watching the MIT lecture about projectile motion and the lecturer asked why $$D=\frac{(v_0)^2\sin(2\alpha)}{g}$$
Why is it $(V_0)^2$ not $V_0$? 
It is a hypothetical question i know that the right answer is $(V_0)^2$.
 A: On an intuitive level, the initial speed $v_0$ can be considered to have two effects: one on the horizontal velocity, and one on the vertical; the former affects the range in a direct sense, and the latter increases the time the projectile is in the air. The combination of both of these gives an overall $v_0^2$ contribution. 
If you were to increase the horizontal velocity on its own (not touching the vertical), you would proportionally increase the range. Similarly if you increased the vertical velocity you would increase the range. Increasing both of these, which is what increasing $v_0$ does, will increase the range twice, but multiplicatively. Think of it like a square - if you increase both sides by the same amount (e.g. 3 times), you increase the area by that amount squared (9 times).
A: Assume that alpha is angle relative to horizontal.
Swap cos and sin below if alpha is angle wrt vertical.  
Time relates to time to rise to "apogee" and return.
Standard expression for time to apogee is 
$t = V/g$   ... 0
This is just stated here but can be derived if desired
Time of flight T relates to initial vertical velocity = $V \times \sin(alpha)$
Total time = ascent + return time = 2 x ascent time.  
$T = 2 * V/g * \sin(\alpha)$ ... 1  = 2 x '0' above  
Horizontal distance D = time x horizontal velocity 
$D = T * V * \cos(\alpha)$ ... 2 
Substitute (1) into (2), replacing T
$D = [2 * V/g * \sin(\alpha)] * [V x \cos(\alpha)]$ ... 3  
$D = 2 * {V^2}/g * \sin(\alpha) * \cos(\alpha)$
This has $V^2$ as in the original.
And $\mathrm{sin}(2\alpha)=2\mathrm{sin}(\alpha)\mathrm{cos}(\alpha)$
A: When a projectile is launched at an initial velocity $V_0$ and an angle $\theta$, the projectile has both a horizontal ($x$-) component and vertical ($y$-) component, as shown below:

From the diagram, you can see that $v_x=v_0 \mathrm{cos\theta}$ and $v_y=v_0 \mathrm{sin\theta}$.
If we try to solve for the horizontal range of your projectile, we can use the equation $$v_x=\frac{\Delta x}{\Delta t}$$ where $\Delta x$ is the horizontal distance traveled by the projectile and $\Delta t$ is the time the projectile spends in air.  If we rewrite $v_x$ in terms of the initial velocity $v_0$ we get that $$v_0 \mathrm{cos\theta}=\frac{\Delta x}{\Delta t}$$. If we clean up the expression a little and solve for the range, $\Delta x$ we get that $$\Delta x= \Delta t v_0 \mathrm{cos\theta}$$.  The problem, though, is that we don't know how long the projectile is in air.  Fortunately, we can determine the time the projectile spends in air by looking just at the vertical motion of of the projectile.  Because the projectile accelerates while in air, due to gravity, the motion in the vertical direction is given by $$\Delta y= \frac{1}{2}g(\Delta t)^2 +v_{0y}\Delta t$$ where $\Delta y$ is the vertical displacement of the projectile, $v_{0y}$ is the initial velocity in the $y$-direction, and $\Delta t$ is the same time interval that we saw in the equation for motion in the $x$-direction. If the projectile is launched on a level surface (which you have assumed, but not stated, in the equation in your question) then $\Delta y=0$.  Additionally, $v_{0y}$ can be given by $v_y=v_0 \mathrm{sin\theta}$ like we saw earlier.  As a result $$0=\frac{1}{2}g(\Delta t)^2+\Delta t v_0 \mathrm{sin\theta}$$  If we try to solve for $\Delta t$, we see that one trivial solution is that $\Delta t=0$.  The other solution is $$\Delta t=\frac{-2v_0\mathrm{sin}\theta}{g}$$  If we substitute $\Delta t$ into our horizontal motion equation, we get that $$\Delta x= \frac{-2v_0^2 \mathrm{sin}\theta \mathrm{cos}\theta}{g}$$  At first this doesn't look like what you have, but there is a trig identity that states that $$\mathrm{sin}(2\theta)=2\mathrm{sin}(\theta)\mathrm{cos}(\theta)$$ so the above expression simplifies to $$\Delta x=\frac{-v_0^2 \mathrm{sin}(2\theta)}{g}$$ where the negative sign is just a consequence of assuming $g=-9.8m/s^2$ so that the negative sign associated with gravity is already tucked inside the symbol.
