# A body thrown vertically upwards takes ‘t’ time to reach the highest point. What would be the time required to reach the same point during its return?

Let $h$ be the height reached by the body in time $t$ when thrown vertically upwards by a velocity $u$. Then, $v=u-gt$ $0=u-gt$ $\therefore t = \frac{u}{g}$

and, $v^2 = u^2 – 2gh$ $0 = u^2 – 2gh$ $\therefore h = \frac{u^2}{2g}$

When the body comes down, it has to travel down the same distance. $\therefore h’ = h = \frac{u^2}{2g}$

The body comes down with initial velocity $u’ = 0$.

$\therefore h’ = u’t+\frac{1}{2}g{t’}^2$

$h’ = 0 + \frac{1}{2}g{t’}^2$

$\therefore t’ = \sqrt{\frac{2h’}{g}} = \sqrt{\frac{2u^2}{g×2g}} = \frac{u}{g} = t$

Thus, the time required to reach the highest point and the time required to come back to the same point is same.

[The equations used above are deduced from Motion Under Gravity]

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