show that time of ascent is equal to time of descent

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show that time of ascent is equal to time of descent

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Use energy at a given height to prove that speed is the same on the way up and on the way down.
Initially the body is thrown vertically so,\[V=u+at\] the final velocity is Zero as it reaches the max. height So,\[t=u/g\] now the body has reached the max. height.. now the body acts as a freely falling body....\[T=\sqrt{2h/g}\] from \[v ^{2}-u ^{2}=2as\]\[H=u ^{2}/2g\] substituting the above in 'T' we get..\[\sqrt{2/g*u ^{2}/2g}\] solving that.... we get\[\sqrt{u ^{2}/g2}\] So\[T=U/g\]....... so time of ascent is equal to time of descent..

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