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

\frac{u}{g}

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u g

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\frac{2 u}{g}

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\frac{u}{2 g}

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Solution

The correct option is B $\frac{2 u}{g}$
$v = u + g t$
at highest distance,v $=$ 0,and g acts in opposite direction.
So,
$0 = u - g t$
$t = \frac{u}{g}$ since time taken to ascent and descent is same.
So, the total time(t) is $\frac{2 u}{g}$ for which it remain in air.

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

The correct option is B 2ugv=u+gtat highest distance,v=0,and g acts in opposite direction.So,0=u−gtt=ug since time taken to ascent and descent is same.So, the total time(t) is 2ug for which it remain in air.

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

The correct option is B $\frac{2 u}{g}$
$v = u + g t$
at highest distance,v $=$ 0,and g acts in opposite direction.
So,
$0 = u - g t$
$t = \frac{u}{g}$ since time taken to ascent and descent is same.
So, the total time(t) is $\frac{2 u}{g}$ for which it remain in air.