Defying Gravity!- 2-Sided

Consider two surfaces parallel to each other in a controlled environment. The environment is made in such a way that the direction of the acceleration due to gravity can be flipped i.e. if the direction of gravity is perpendicular to one of the surfaces, the direction can be switched such that it is perpendicular to the other surface. Now, a ball is dropped from one of the surfaces, and falls to the other. However, after a certain time, the direction of acceleration is flipped, such that when the ball reaches the other surface, its velocity is zero. Then after every time period T T , the gravity is flipped such that the ball continues to reach the surfaces with a velocity equal to zero. If the distance between the two surfaces is 10 10 m . m. and the acceleration due to gravity in the system is 10 10 m s 2 ms^{-2} , find time period T T .

Also, after solving, generalize a formula for the time period T T for the surfaces at distance k k apart, with acceleration due to gravity as g g

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The answer is 2.

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1 solution

Kashish Gupta
Jun 22, 2014

By symmetry we can say that that point at which gravity must be flipped is midpoint. So 5=1/2gt^2 g=10 given, so t=1 So at first gravity must be flipped after 1second and after that every 2t second I.e. 2seconds answer

Can you derive the expression for time period T T , when the distance between the two surfaces is k k , and a a due to gravity is g g ?

Nanayaranaraknas Vahdam - 6 years, 11 months ago

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Here is it!

T = 2 k g T=2\sqrt{\frac{k}{g}}

Muhammad Arifur Rahman - 5 years, 8 months ago

t = 2 g k t=2\sqrt { \frac { g }{ k } }

Ronak Agarwal - 6 years, 11 months ago

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