Question

The time of ascent of a body projected with a velocity $u\,m{s^{ - 1}}$ is _____:
A. $g/u$
B. ${u^2}/2g$
C. $u/g$
D. $2u/g$

Answer 1

To solve this question, with the help of velocity of projection and the angle of projectile, we can find the time of flight. After finding the time of flight, as we know, time of ascent is equal to the half of the time of flight. Now, we can find the time of ascent.

Complete step by step solution:
According to the question, when body is projected vertically upwards, angle of projection:
$\theta = {90^\circ }$ .
Velocity of projection is $u$ (given).
So, time of flight:
$T = \dfrac{{2u.\sin \theta }}{g}$
$\begin{gathered} \Rightarrow T = \dfrac{{2u.\sin {{90}^\circ }}}{g} \\\ \Rightarrow T = \dfrac{{2u.1}}{g} = \dfrac{{2u}}{g} \\\ \end{gathered}$
Since, the 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, $t = \dfrac{T}{2} = \dfrac{u}{g}$
Hence, the correct option is (C.) $u/g$ .

Note:
The time of flight of a projectile motion is the time from when the object is projected to the time it reaches the surface. Once we know the equation of the time of flight, by putting a dirty equation value in the displacement of the body along the x-axis, we can calculate the total distance travelled by the body along the x-axis.