I have been looking at this question from a Mechanics Chapter:-(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Motion of a body under acceleration of gravity

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