Bolt dropped from Moon Skyscraper!!!

Classical Mechanics Level 3

Imagine, we've colonized on the Moon! We're building huge Skyscrapers. And its easier in the Moon's low gravity than Earth's.

An engineer, at the top of a nearly finished Skyscraper, trying to join and tighten the last bolt of the building. But unfortunately he drops it!

Then what was the total time of flight, T for the bolt?

Details and Assumptions:

  1. The height of the Skyscraper is H .

  2. As there's no atmosphere in the Moon, air can't affect the motion of the bolt. So you can consider the bolt as a Freely Falling Body From Rest .

  3. The height of each floor is 1 n H \frac{1}{n}H .

  4. The bolt took t seconds to cross the Ground Floor .

  5. Acceleration due to gravity, g = 10 m s − 2 10ms^{-2}

  6. Moon's gravity is 1 6 \frac{1}{6} of Earth's.

......Hope you don't make a Spontaneous or Random choice......!!!

( n − 1 ) t n + ( n − 1 ) n + 1 \frac{(n-1)t}{\sqrt{n}+(n-1)\sqrt{n+1}} n t n − n − 1 \frac{\sqrt{n}t}{\sqrt{n}-\sqrt{n-1}} n t n − n − 1 \frac{nt}{n-\sqrt{n-1}} 2 n t n − n − 1 \frac{\sqrt{2}nt}{n-\sqrt{n-1}}

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Muhammad Arifur Rahman by Muhammad Arifur Rahman , 23, Bangladesh

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1 solution

Let's analyze the Clues given:

  1. The height of the Skyscraper is H . So you can consider the bolt as an object falling from height, H (and from Rest , of course).

  2. The height of each floor is 1 n H \frac{1}{n}H . So height of the Ground Floor is the same!

  3. The bolt took t seconds to cross the Ground Floor.

This Necessarily means that The bolt crossed 1 n H = H n \frac{1}{n}H=\frac{H}{n} in the last t seconds .

Now let's make some equations:

  1. Assume, total time of flight= T The bolt crossed H distance in this time.

    Assume,the acceleration due to gravity for Moon= g

    So H = u t + 1 2 g T 2 H=ut+\frac{1}{2}gT^2 → H = 1 2 g T 2 \rightarrow H=\frac{1}{2}gT^2

  2. The bolt crossed H − H n = H ( 1 − 1 n ) H-\frac{H}{n}=H(1-\frac{1}{n}) distance in ( T − t ) (T-t) seconds [Note this! You must have to understand it.] .

So, H ( 1 − 1 n ) = 1 2 g ( T − t ) 2 H(1-\frac{1}{n})=\frac{1}{2}g(T-t)^2 → H = 1 2 g ( T − t ) 2 1 − 1 n \rightarrow H=\frac{\frac{1}{2}g(T-t)^2}{1-\frac{1}{n}}

From the above two equation, let's do some little Algebra.

I will briefly describe every step, for easy understanding. You may shorten it as you need!

1 2 g T 2 = 1 2 g ( T − t ) 2 1 − 1 n \frac{1}{2}gT^2=\frac{\frac{1}{2}g(T-t)^2}{1-\frac{1}{n}}

→ 1 2 g T 2 = 1 2 n g ( T − t ) 2 ( n − 1 ) \rightarrow\frac{1}{2}gT^2=\frac{1}{2}\frac{ng{(T-t)}^2}{(n-1)}

→ T 2 = n ( T − t ) 2 ( n − 1 ) \rightarrow T^2=\frac{n{(T-t)}^2}{(n-1)}

→ T = n ( T − t ) n − 1 \rightarrow T=\frac{\sqrt{n}(T-t)}{\sqrt{n-1}}

→ T ( n − 1 ) = n T − n t \rightarrow T(\sqrt{n-1})=\sqrt{n}T-\sqrt{n}t

→ T ( n − n − 1 ) = n t \rightarrow T(\sqrt{n}-\sqrt{n-1})=\sqrt{n}t

→ T = n t n − n − 1 \rightarrow T=\frac{\sqrt{n}t}{\sqrt{n}-\sqrt{n-1}} Yahoo! Its the answer!!!

You may notice that the equation doesn't contain g at all. Which means the time T is *g Independent * .

So no matter whether its Earth,Moon or not. The same equation is applicable for Mars, even for Comet 67-P! Maybe this equation will be helpful for our *Mars Colonization * "!!!

Photo Credit: Red Orbit

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