From which equation of motion was this formula derived from in physics

Question

When solving problems involving projectile motion I use:

$\sqrt{2 * \dfrac{\text{height above ground}}{9.8}}$

Eg calculate the time it takes for a bomb to impact if it is travelling 4.9km above ground I simply do:

$\sqrt{2 * \dfrac{49000}{9.8}}$ and get the correct answer of 100 seconds.

However, from traditional equations of projectile motion ( Which I am required to use, and not this formula) I have to use either one of these which relate to the vertical velocity of the object:

So my question is the above formula mentioned derived from the above 3 ?

Answer 1

The 2nd one. Solve for y=height and $u_0=0$.

$V^2 = 2 a h$ , $a=g=9.8$

h=$\sqrt{(v^2/2g)}$

The. Plug this into eother eq1 or 3.

Answer 2

If you use the equation

$$\Delta y = u_yt + \frac{1}{2}a_yt^2$$

with $u_y = 0$ and $a_y = g$, you obtain

\begin{align*} \Delta y & = \frac{1}{2}gt^2\\ 2\Delta y & = gt^2\\ \frac{2 \Delta y}{g} & = t^2\\ \sqrt{\frac{2 \Delta y}{g}} & = t \end{align*}

Substituting $-4900~\text{m}$ for $\Delta y$ and $-9.8~\dfrac{\text{m}}{\text{s}^2}$ for $g$ yields

$$t = \sqrt{\frac{2\Delta y}{g}} = \sqrt{\frac{2 \cdot -4900~\text{m}}{-9.8~\frac{\text{m}}{\text{s}^2}}} = \sqrt{1000~\text{s}^2}$$

where I have used the fact that $$4.9~\text{km} \cdot \frac{1000~\text{m}}{\text{km}} = 4900~\text{m}$$