Prove that time of ascend= time of desend (numerically)

1.	Prove that time of ascend= time of desend (numerically)
Answer» Thanku Time of Ascent:The time taken by body thrown up to reach maximum height h is called as time of ascent.Let t₁ be time of ascent.at maximim height , its velocity is zero, V=0hence , V=u-gt0=u-gt₁t1=u/g ------------- (1)Time of descent:after reaching the maximum height, the body begins to travel downwards and reaches the ground.The time taken to by a freely falling body to touch the ground is called time of descent.Initial velocity for downward journey is zero , u= 0let t₂ be time of descent.s=ut+1/2 gt²h=0+1/2gt₂²t2=√2h/gbut, h=u²/2gso, we gett₂=√2u²/2g =u/g---------------(2)from equation 1 and 2 we can sayt₁=t₂the time of ascent is equal to time descent in the case of bodies moving under gravity.

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