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

\frac{g}{u}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

u^{2} / 2 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{u}{g}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2 u}{g}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{u}{g}$
Taking upward direction to be positive.
Initial velocity is $u$ .
Final velocity at the maximum height $v = 0$
Acceleration $a = - g$
Using $v = u + a t$
$∴$ $0 = u - g t$
$⟹ t = \frac{u}{g}$

Suggest Corrections

Similar questions

Q. Question 2
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a)

\frac{u}{2}

(b)

\frac{u^{2}}{2 g}

(c)

\frac{u^{2}}{g}

(d)

\frac{u}{2 g}

Q. The time of ascent of a body projected with a velocity

{u m s}^{- 1}

is _____ .

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ . [ $g$ is the acceleration due to gravity]

A body is projected upward with a velocity u. It passes through a certain point above the ground after $t_{1}$ . From this instant, the time after which the body passes through the same point during the return journey is

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ :

Join BYJU'S Learning Program

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

The correct option is C ugTaking upward direction to be positive.Initial velocity is u.Final velocity at the maximum height v=0Acceleration a=−gUsing v=u+at∴ 0=u−gt⟹ t=ug

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

The correct option is C $\frac{u}{g}$
Taking upward direction to be positive.
Initial velocity is $u$ .
Final velocity at the maximum height $v = 0$
Acceleration $a = - g$
Using $v = u + a t$
$∴$ $0 = u - g t$
$⟹ t = \frac{u}{g}$