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A,B,C and D are four points on a vertical line such that AB = BC CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, CD should be in the ratio 1 : (sqrt(2)-1) : (sqrt(3)-sqrt(2)). |
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Answer» Solution :Let AB = BC = CD = x and time taken by the body to cover these distances be`t_(1), t_(2)`and `t_(3)` respectively. Now, `x = (1)/(2)"gt"_(1)^(2) """or", t_(1) = sqrt((2x)/(G)) ""cdots(1)` `2x = (1)/(2) g(t_(1)-t_(2))^(2) """or", t_(1)+t_(2)= sqrt((4X)/(g)) ""cdots(2)` `3x = (1)/(2) g(t_(1)+t_(2)+t_(3))^(2)"""or" t_(1)+t_(2)+t_(3)= sqrt((6x)/(g)) "" cdots(3)` From (1) and (2) we GET, `t_(2) = sqrt((4x)/(g))-sqrt((2x)/g)=sqrt((2x)/(g))(sqrt(2)-1)` From (2) and (3) we get , `t_(3)=sqrt((6x)/(g))-sqrt((4x)/g)=sqrt((2x)/(g))(sqrt(3)-sqrt(2))` `:. "" t_(1):t_(2):t_(3)=1:(sqrt2-1):(sqrt3-sqrt(2))""`(Proved). |
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