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A particle is dropped from the top a tower `h` metre high and at the same moment another particle is projected upward from the bottom. They meet the upper one has descended a distance `h//n` . Show that thevelocities of the two when they meet are in the ratio `2`: `(n-2)` and that the initial velocity of the particle projected upis `sqrt((1//2))n gh`.

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Correct Answer - `(2)/(n-2)`
`t =sqrt((2h)/(ng)), h-(h)/(n)=ut-(1)/(2) g t^(2)`
`rArr u=(h)/(t) =sqrt((nhg)/(2))`
`v_(1)=g t, v_(2)=u-g t, (v_(2))/(v_(1)) =(u-g t)/(g t) =(u)/(g t)-1`
`rArr (v_(2))/(v_(1)) =(h)/(g t^(2) )-1 =(hng)/(g2h)-1 rArr (v_(1))/(v_(2))=(2)/(n-2). `
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