I have been looking at this question from a Mechanics Chapter:- A stone is thrown vertically upwards with a velocity of u metres per second. It passes a ledge in t1 seconds and repasses it t2 seconds after the start. Find the height of the ledge? Take g = 9.8 metres per second sq This is the question in entirety. I have tried using s = ut + 1/2 a t^2 substituting g in as s = ut - 1/2 g t^2 If the height is the same at both these times then I have assumed that:- ut1 -1/2 g t1^2 = ut2 - 1/2 g t2^2 Then factoring out t1 and t2 t1(u - 1/2 g t1) = t2(u - 1/2 g t2) then failed to see how this could find s! Then tried rearranging the original formula in terms of u and t1, so: u = (s + 1/2 g t1^2)/t1 and substituting into s = ut2 - 1/2 g t2^2 but couldn't proceed. Can somebody help? I am assuming that I need to find s in terms of g, t1 and t2 rather than an actual value.