If a ball is thrown up, is the time taken by it for the upward and downward motion equal? Explain.

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Solution

Let the ball be projected upwards with an initial velocity $v_{o}$ .
For the upward journey,
Initial velocity, $u = v_{o}$
Final velocity, $v = 0$
Acceleration, $a = g$ (in downward direction)
From the first equation of motion, $v = u + a t$
$\Rightarrow 0 = v_{o} - g t_{u p}$
$\Rightarrow t_{u p} = \frac{v_{o}}{g}$
Maximum height reached, $h = \frac{{v_{o}}^{2}}{2 g}$
For the downward journey,
Initial velocity, $u = 0$
Distance moved, $h = \frac{{v_{o}}^{2}}{2 g}$
From the second equation of motion, $h = u t + \frac{1}{2} g t^{2}$
$\Rightarrow \frac{{v_{o}}^{2}}{2 g} = \frac{1}{2} g {t_{d o w n}}^{2}$
$t_{d o w n} = \frac{v_{o}}{g}$
Hence, $t_{u p} = t_{d o w n}$

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