A body is projected vertically upwards with an initial velocity $u$ . If the acceleration due to gravity is $g$ , find the time wherein it remains in the air.

\frac{u}{g}

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\frac{2 u}{g}

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\frac{3 u}{g}

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\frac{4 u}{g}

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Solution

The correct option is B $\frac{2 u}{g}$
Given
Initial velocity $= u$
Acceleration, $a = - g$ (direction of g is opposite to that of u)
Clearly, the body would remain in the air until the time when it returns to the ground.
Therefore, displacement, $s = 0$
From the equation of motion,
$s = ut + \frac{1}{2} {at}^{2}$
Putting the values, we get:
$0 = ut - \frac{1}{2} {gt}^{2}$
$t (gt - 2 u) = 0$

$t = 0$ or $t = \frac{2 u}{g}$

Since $t = 0$ represents the initial state of the body, therefore $t = \frac{2 u}{g}$ .

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A body is projected vertically upwards with an initial velocity u. If the acceleration due to gravity is g, find the time wherein it remains in the air.

A body is projected vertically upwards with an initial velocity $u$ . If the acceleration due to gravity is $g$ , find the time wherein it remains in the air.