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Classical Mechanics Level 2

Two bodies of different masses M and m are dropped from two different heights a and b. The ratio of times taken by the two bodies to drop through these distances is :

Note- Please do post your solutions!

a^{2} : b^{2} a : b \sqrt{a} : \sqrt{b} \frac{M x b}{m x a}

Archiet Dev by Archiet Dev , 20, India

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2 solutions

Micah Wood
Oct 26, 2014

Assume that there is no air resistance present, formula for time t t taken for an object to fall distance d d is t = 2 d g t=\sqrt{\frac{2d}{g}}

So for height a a , we have t a = 2 a g t_a=\sqrt{\frac{2a}{g}} and for height b, t b = 2 b g t_b=\sqrt{\frac{2b}{g}}

So ratio of times taken by the two bodies to drop through these distances is: t a : t b = 2 a g : 2 b g = 2 g a : 2 g b = a : b \begin{array}{c}~t_a:t_b &= \sqrt{\dfrac{2a}{g}}:\sqrt{\dfrac{2b}{g}}\\~\\&=\sqrt{\dfrac{2}{g}}\sqrt{a}:\sqrt{\dfrac{2}{g}}\sqrt{b}\\~\\&=\boxed{\sqrt{a}:\sqrt{b}} \end{array}

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Bala Murugan
Nov 7, 2014

i considered this way.if M is greater then it sholud be dropped from a little higher.as mass increases its velocity increases.so their velocity has to be made srqt of distance .this becomes different if there is no air

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