neutrino_super =>can you solve this ?????????????????????????????????????????????

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

Ma/Mb x b/a a^2 : b^2 a^1/2 : b^1/2 a : b

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1 solution

Dawar Husain
Dec 22, 2014

a = u t + a t 1 2 2 = g t 1 2 2 a=ut+\dfrac{at_{1}^{2}}{2} = \dfrac{gt_{1}^{2}}{2}

t 1 = 2 a g \ \ \ \implies t_1 = \sqrt{\dfrac{2a}{g}}

b = u t + a t 2 2 2 = g t 2 2 2 b=ut+\dfrac{at_{2}^{2}}{2} = \dfrac{gt_{2}^{2}}{2}

t 2 = 2 b g \ \ \ \implies t_2 = \sqrt{\dfrac{2b}{g}}

t 1 t 2 = 2 a 2 b \dfrac{t_1}{t_2}=\dfrac{\sqrt{2a}}{\sqrt{2b}}

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