The time of ascent of a body projected with a velocity ${u m s}^{- 1}$ is _____ .

g/u

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u^{2} / 2 g

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u/g

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2u/g

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Solution

The correct option is C u/g
When body is projected vertically upwards, angle projection $θ = 90^{o}$
Velocity of projection is $u$ (Given)
So, time of flight $T = \frac{2 u sin θ}{g}$
$⟹ T = \frac{2 u sin 90}{g} = \frac{2 u}{g}$
Since time of ascent is equal to the time of descent. Thus, time of ascent is equal to the half of the time of flight.
Time of ascent $t = \frac{T}{2} = \frac{u}{g}$

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The time of ascent of a body projected with a velocity u ms−1 is _____ .

The time of ascent of a body projected with a velocity ${u m s}^{- 1}$ is _____ .