An object is dropped from a certain point A at a height 'h' from the ground. During it's journey straight downwards, the object passes points B and C such that the ratio of time taken t₁ to cover AB and t₂ to cover BC is 1: ( 2 - 1) . What is the ratio of distances AB:BC ?

Step-by-step solution

Let the time taken to travel from A to B be t₁ = t . The time taken to travel from B to C is t₂ = ( 2 - 1)t . The total time taken to travel from A to C is t_ AC = t₁ + t₂ = t + ( 2 - 1)t = 2 t . Since the object is dropped from rest at A, the initial velocity is zero. The distance AB is given by AB = 1 2 gt₁^2 = 1 2 gt^2 . The distance AC is given by AC = 1 2 gt_ AC ^2 = 1 2 g( 2 t)^2 = gt^2 . The distance BC is BC = AC - AB = gt^2 - 1 2 gt^2 = 1 2 gt^2 . The ratio of distances AB and BC is AB : BC = 1 2 gt^2 : 1