Intuitive meaning of factor 2 in formula of vertical throw max height $h=v^2/2g$ This is a question about a simple thing. The simplified expression for maximum height in vertical throw is $h=\frac{v^2}{2g}$ , could anyone explain intuitively (analogies are welcome) why there is a factor 2 on the denominator? And I mean intuitively, so not by transforming formulas etc, but by saying what that constant actually represents in this model.
 A: The factor $2$ comes from the equation $v_f^2 = v_i^2 - \color{#C00}2gh$. That's how we derive the equation for max height.$$v_f^2 = v_i^2 - 2gh$$Now, we want $v_f$ to be $0$ (because at the maximum height, the speed is always zero). Calculating $h$, we have:$$ 0 =  v^2 - 2gh \Rightarrow -v^2 = -2gh \Rightarrow v^2 = 2gh \Rightarrow h = \dfrac{v^2}{2g} $$You could skip the actual derivation because the intuition is in the first sentence.
A: The height is equal to the average velocity times the time to reach the maximum height, so $$h = v_{ave} t.$$
The acceleration due to gravity is the change in velocity divided by time, so by rearranging $$t =(v_{final} - v_{initial})/(-g)=(0-v)/(-g)=v/g. $$
Now, remembering that for uniformly accelerated motion the average velocity is half the sum of the initial and final velocities, $$v_{ave} = (0+v)/2=v/2.$$
Then by combining those three equations, $$h=(v/2)(v/g)=v^2/2g.$$
Essentially the factor of one-half comes from the averaging of the variable velocity over the trajectory. 
A: The best intuition is a calculation but in this simple case, the calculation is really intuitive so you shouldn't turn off when you hear the word "calculation".
The height reached by initial velocity $v$ is the height of the object after the initial velocity $v$ drops to $0$ (and then reverts the sign) because of the downward acceleration $g$.
How much time does it take to reduce the velocity from $v$ to $0$? Well, the acceleration is the velocity change per unit time and it is $g$. So the time needed to reduce the velocity to zero is $$t=\frac vg.$$  
Now, how  far the object gets after time $t$? It's simple: the distance is the velocity times time. But the velocity is changing. You must use the average velocity to calculate the distance: 
$$h=\bar v\cdot t.$$
What is the average velocity? Because the velocity is dropping linearly, it's just $1/2$ of the sum of the initial velocity $v$ and the final velocity $0$:
$$\bar v = \frac{v+0}{2} = \frac v2 $$
So we have 
$$h=\bar v\cdot t = \frac v2\cdot t = \frac v2\cdot \frac vg = \frac{v^2}{2g}$$
The factor of $1/2$ came from the need to calculate the average velocity and the average of two numbers is one-half ($1/2$) of their sum.
I don't need any kinetic energy to do the calculation, as you can see. But the factor $1/2$ in the formula for the kinetic energy $E=mv^2/2$ has a totally analogous origin: it is the total work you have to do to accelerate the object from velocity $v=0$ to velocity $v$. The force one has to overcome is $F=ma$ and the work is $F\cdot \Delta x$ but $\Delta x=\bar v \cdot t = (0+v)t/2$ so we have $$ E = ma\cdot x = ma\cdot vt/2 = mv^2/2$$ where I used $v=at$ because the velocity was assumed to increase uniformly again. Once again, I calculated the average velocity which turned out to be $1/2$ of the maximal (final) one.
Equivalently, the factor $1/2$ comes from the fact that the indefinite integral of $x$ over $x$ is $x^2/2$. For example, the integral $\int v\,dt = \int at\,dt=at^2/2$ gives the total distance travelled in accelerated motion – and all the other examples above are analogous. But one doesn't have to learn the full-fledged theory of integration for this outcome: the integral is the area under the curve and for a linear function $y=x$ between $x=0$ and $x=x$, the area is just a triangle whose area is simply $1/2$ of the area of the square or rectangle; again, this $1/2$ is the same $1/2$ we obtained from the averaging above. So the consequences of this particular simple integral may be understood even without any broader knowledge of integration (or differentiation), and that's what was done above.
A: What is the effect of g, it amounts to the velocity added or subtracted each second from your initial velocity u. From this you quite intuitively know that throwing a ball upwards you will have 0 velocity within $\frac{u}{g}$ seconds (that is you have reached maximum height). Drawing a v-t graph for the given condition:

the area under the v-t curve gives you the displacement, so calculating the area under the graph you get:
$$\frac{1}{2} \times base \times height = \frac{1}{2} \times \frac{u}{g} \times u = \frac{u^2}{2g}$$
seems quite intuitive to me this way.
So answering your question why there is a factor of $\frac{1}{2}$, you could think of it as $\frac{u^2}{g}$ is the total distance traveled (twice the maximum height) but you only need half the total distance traveled.
