A, B, C are points in a vertical line such that AB = BC. If a body falls freely from rest at A, and tj and t2 are times taken to travel distances AB and BC, then ratio t1/t2 is
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Explanation:
Let ab = bc = cd = x
=> x = (1/2)gt1^2
2x = (1/2)gt2^2
3x = (1/2)gt3^2
=>
t1 = √(2x/g)
t2 = √(4x/g) and
t3 = √(6x/g)
=>
ratio of times of descent through ab, bc and cd
= t1 : (t2 - t1) : (t3 - t2)
= √(2x/g) : √(4x/g) - √(2x/g) : √(6x/g) - √(4x/g)
= √2 : (2 - √2) : (√6 - 2).
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