Question

The time of ascent of a body projected with a velocity $u$ ${m{s^{-1}}}$ is ………………
A) $\dfrac{g}{u}$
B) $\dfrac{{{u^2}}}{{2g}}$
C) $\dfrac{u}{g}$
D) $\dfrac{{2u}}{g}$

Answer 1

Hint: We know that anybody projected into the air at angle other than ${90^0}$ with the horizontal near the surface of the earth, is called a projectile.
Time of ascent is the time taken by the body to reach the maximum height from the initial position.

Complete step-by-step answer:
For a projectile motion, the time to reach maximum height is called time of ascent.
Examples for projectile motion are a cricket ball thrown by a fielder, a bullet fired from a gun.
For a projectile, the vertical component velocity ${v_y}$ is zero at the highest point.

In the diagram velocity vector is resolved into two components. One is along the x axis that is ${u_x} = u\cos \theta $ and one more along y axis that is ${u_y} = u\sin \theta $.
During time of ascent motion is along y axis. Therefore consider the component of y axis in the below equation.
We have, ${v_y} = {u_y} + {a_y}t$
we have ${u_y} = u\sin \theta $
In vertical direction, the acceleration of the projectile is equal to the free fall acceleration which is constant and always directed downward. ${a_y} = - g$
${v_y} = u\sin \theta - gt$
Here, at maximum height final velocity is zero.${v_y} = 0{\text{ and }}t = {t_a}$
$\therefore {t_a} = \dfrac{{u\sin \theta }}{g}$
Therefore,${t_a} = \dfrac{{{u_y}}}{g} = \dfrac{u}{g}$

Option C is the correct answer.

Additional information:
Let a body be projected at O with an initial velocity u that makes an angle $\theta $ with the X-axis.
Due to the fact that two dimensional motions can be treated as two independent rectilinear motions, the projectile motion can be broken up into two separated straight line motions.
Horizontal motion with zero acceleration (that is constant velocity as there is no force in horizontal direction).
 Vertical motion is having a constant downward acceleration = g ( $\because $ it is moving under gravity).

Let the body reach a point P(x,y) in its trajectory after time t. that is, horizontal displacement and vertical displacement of the body in time t are X and Y respectively.
Let us consider the horizontal motion. As the horizontal motion has no acceleration the horizontal component of projectile’s velocity ${u_x}$ remains constant throughout the motion, the displacement of the projectile after any time ‘t’ from the initial position is given by $x = {u_x}t + \dfrac{1}{2}{a_x}{t^2}$ . Since horizontal acceleration is zero. $x = {u_x}t = \left( {u\cos \theta } \right)t$
Now let us consider the vertical motion. In vertical direction, the acceleration of the projectile is equal to the free fall acceleration which is constant and always directed downward. ${a_y} = - g$ .

Note: In this type of question, remember if a body moving upward, acceleration due to gravity acting downward then, a=-g . Similarly, if a body is coming downward then a=+g.
As the horizontal motion has no acceleration the horizontal component of the projectile's velocity ${u_x}$ remains constant throughout the motion.