A body is projected vertically upward with an initial velocity $u .$ If acceleration due to gravity is $g,$ time for which it remains in air is:

$\frac{u}{g}$

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$\frac{2 u}{g}$

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$u g$

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$\frac{g}{2 u}$

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Solution

The correct option is B
$\frac{2 u}{g}$

When body reaches highest point, final velocity, $v = 0$
Let the time required to reach highest point be $t_{1}$
Using equation of motion:
$v = u + a t$ , we get
$v = u - g t_{1}$ [as $a = - g$ ]
$\Rightarrow 0 = u - g t_{1}$
$\Rightarrow g t_{1} = u$
$\Rightarrow t_{1} = \frac{u}{g}$
Time taken by the body to reach the highest point $=$ time taken by body to come back to the initial point after reaching the highest point.
Therefore,
Total time $= 2 \times$ Time taken to reach highest point
Hence, total time $= \frac{2 u}{g}$

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A body is projected vertically upward with an initial velocity u. If acceleration due to gravity is g, time for which it remains in air is:

A body is projected vertically upward with an initial velocity $u .$ If acceleration due to gravity is $g,$ time for which it remains in air is: