Let us consider a man who is jumping from a 5-storey building on Earth dies as soon as he reaches the surface. From which floor (minimum) must he be jumped in order to die on the Moon?
Consider:
1) Man's mass = 70 kg
2) Initial velocity = 0 m/s
3) Earth's gravity = 9.8 m/s
4) Moon's gravity = 1.6 m/s
5) Height of each floor = 3 m
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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:
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 $\left|v_{\text{Moon}}\right| = \left|v_{\text{Earth}}\right|$ .
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 $\left( \dfrac{H}{3} \right) = \left \lceil 30.625 \right \rceil = \boxed{31}$
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Yes you are correct , if we use $g_{\text{moon}}=\dfrac{g_{\text{earth}}}{6}$ we will indeed get a different answer from the original one $g_{\text{moon}}=1.6ms^{-2}$ .... i hope Sai Krishna Dittakavi will take the necessary steps to solve this problem.
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I'm sorry, I thought you had posted the problem.
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!
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But it is misleading. You should've actually stated that acceleration due to gravity on the moon is $1.633...$ or simply $1.63$ would have been okay. I typed in 31 at first and then I realized what assumption you must've made so I typed in 30.
a= $\frac{1}{6}$ g where a= gravitational acceleration on moon
5 h g=x h a
x=30 :v easy
very very easy
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$\text{Velocity attained just before hitting the ground must be same in both the conditions (earth and moon) } \\ \text{Let height of each Floor be } h \text{ m and Using } v^2-u^2=2as \\ v_{e} = v_{m} \implies \sqrt{2\cdot g(5h)} = \sqrt{2\cdot \dfrac{g}{6}(H)} \\ \dfrac{H}{6}=5h \implies H=30h \\ \text{A 30 storey building is required}$
$\text{None of the information given is required except for } u=0$