Sharky's experience on Metazhaghu

Classical Mechanics Level 4

Sharky the famous astronaut was once sent into space (of course, in a spaceship), in search of a planet that is potentially habitable. However, due to some complication, Sharky accidentally lands on a planet called "Metazhaghu". This planet was a bit strange because the gravitational acceleration there was not constant. Sharky stood on a cliff and dropped a substance and found that the gravitational acceleration $g$ becomes $2g,3g,4g , \ldots$ for equal time intervals of $t$ .

Now Sharky from a certain height drops an object and that object hits the surface (ground) of Metazhaghu after such $n$ time intervals. Find the height from ground of the place from where Sharky dropped the object.

\dfrac{3gt^2n(n+1)(2n+1)}{8}

\dfrac{7gt^2n(n+1)(2n+1)}{6}

\dfrac{gt^2n(n+1)(2n+1)}{12}

\dfrac{7gt^2n(n+1)(2n+1)}{8}

\dfrac{gt^2n(n+1)(2n+1)}{6}

\dfrac{5gt^2n(n+1)(2n+1)}{6}

2 solutions

Nihar Mahajan
Jun 2, 2015

$\large{S_1= (0)t + \dfrac{1}{2}gt^2 = \dfrac{(1)gt^2}{2} \quad v_1 = 0 + gt = gt \\ S_2= (gt)t + \dfrac{1}{2}(2g)t^2 = \dfrac{(4)gt^2}{2} \quad v_2 = gt + 2gt = 3gt \\ S_3= (3gt)t + \dfrac{1}{2}(3g)t^2 = \dfrac{(9)gt^2}{2} \quad v_3 = 3gt + 3gt = 6gt \\ \dots \\ \dots\dots\dots \quad\quad \quad\quad\quad\quad\quad v_{n-1} = \dfrac{n(n-1)}{2}\\ S_n= \left(\dfrac{n(n-1)}{2}gt\right)t + \dfrac{1}{2}(ng)t^2 = \dfrac{(n^2)gt^2}{2} }$

$\Large {\text{Required height} = \sum _{r=1}^n S_r \\ = gt^2\left(\dfrac{1+4+9+16 +\dots + n}{2}\right) \\ = \dfrac{gt^2n(n+1)(2n+1)}{12}}$

I guess there is a small correction, summation of squares of $n$ natural numbers is $= \dfrac{n(n+1)(2n+1)}{6}$ and hence the last step becomes,

$= \dfrac{\dfrac{gt^2n(n+1)(2n+1)}{6}}{2}\\ = \dfrac{gt^2n(n+1)(2n+1)}{12}$

EDIT: Lot's of changes in numerators and denominators. Sorry for inconvenience! $\huge\ddot\smile$

Sravanth C. - 6 years ago

Hi!

How did you get the last step from the 2nd last step ? There's a small typo .

A Former Brilliant Member - 6 years ago

Whoa! Thanks for modifying that!(I didn't know that mod's had these many powers!)

Sravanth C. - 6 years ago

@Sravanth C. – I think he's saying that how did $6$ came in numerator in your $2^{nd}$ last step.

Nihar Mahajan - 6 years ago

@Nihar Mahajan – Sorry, I hit the preview button earlier instead of posting that. :P

Sravanth C. - 6 years ago

Hey Sravanth,sorry to say,but there is still one typo in your comment,the $2$ in the numerator's denominator should be $6$ ,kindly correct it.

Adarsh Kumar - 6 years ago

Why feel sorry?

Sravanth C. - 6 years ago

Yeah , it was a typo.

Nihar Mahajan - 6 years ago

Anyways it was a great question! With lots of creativity . . . $\huge\ddot\smile$

Sravanth C. - 6 years ago

Rohit Gupta
Jun 7, 2015

The same question can be done using the graphs as well.. Draw graph of velocity vs time and find the area under the graph will give the displacement and in calculation of the area we reach to the same series..!!

Sharky's experience on Metazhaghu

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