5 = ?

Thank you. But I found that the information is required for some other ways of solution to the problem... Because, though the answer is unique, it needn't have only one way to approach, always. Right?

Suppose I don't know that the value of acceleration due to gravity on the Moon is ${1}/{6}~th$ of that of Earth. I just consider the values that you've provided to me, which are $g_{\text{Earth}} = 9.8~ m s^{-2}$ and $g_{\text{Moon}} = 1.6~ m s^{-2}$ (Please make that correction too that it is $m s^{-2}$ not $m s^{-1}$ ).

Then, I would approach the problem as:

• On Earth:

Using the equation $v^2 = u^2 -2gh$ , we have,

$\begin{aligned} v^2 & = & u^2 -2gh \\ v^2 & = & 0 - 2 \times (-9.8) \times (5 \times 3) \\ v^2 & = & 294 \\ v & = & \sqrt{294}~ m s^{-1} \end{aligned}$

If he has to die than his velocity before hitting the ground on the Moon must be greater than or equal to his the velocity with which he hits the ground on Earth, that is,

$\left|v_{\text{Moon}}\right| \ge \left|v_{\text{Earth}}\right| \implies v_{\text{Moon}}^{2} \ge v_{\text{Earth}}^{2}$

Without loss of generality let's take $\left|v_{\text{Moon}}\right| = \left|v_{\text{Earth}}\right|$ .

• On Moon:

Again using $v^2 = u^2 -2gh$ , we have,

$\begin{aligned} v^2 & = & u^2 -2gh \\ 294 & = & 0 - 2 \times (-1.6) \times H \\ H & = & \dfrac{294}{3.2} \\ H & = & 91.875 \end{aligned}$

We know that height of each floor is still $3 ~ m$ .

So, required floor $= \dfrac{91.875}{3} = 30.625$ .

The answer comes out not to be a perfect integer. But based upon our conclusion above that if the man needs to die then, $v_{\text{Moon}}^{2} \ge v_{\text{Earth}}^{2}$ .

So,

$3.2H = v^2 > 294$

$\implies H = \dfrac{v^2}{3.2} > \dfrac{294}{3.2}$

$\implies H > 91.875$

$\implies \dfrac{H}{3} > \dfrac{91.875}{3} = 30.625$

Hence, minimum required floor $\left( \dfrac{H}{3} \right) = \left \lceil 30.625 \right \rceil = \boxed{31}$

I agree with all you said. But I have asked you to assume the parameters that I gave you along with the question. Thank you!