Question

What is the time of descent for a body projected vertically upward with velocity $v$?
(A) Same as time of ascent
(B) $\dfrac{v}{g}$
(C) $\dfrac{u}{g}$ , $u$ = final velocity
(D) Both A and B

Answer 1

Hint
When the ball will be thrown upward it will undergo retardation due to gravity g. During the ascent the final velocity of the ball will become zero and then by using Newton’s law of equation we get time of ascent. And for descent the body will undergo acceleration and then by again using Newton's law of equation we get the time of descent.

Complete step by step solution
Let us consider a body is projected upward with velocity $v$.
During ascent the body will feel retardation due to gravity $g$, let the time of ascent is $t_a$ and in the last seconds the final velocity of the body will become to zero. Now, using Newton’s law of equation, we get
$v = u - g{t_a}$ , $v$ is the final velocity during ascent and $u$ is the initial velocity.
As, $v$ = 0 for ascent then above equation will become
$\Rightarrow 0 = u - g{t_a}$
$\Rightarrow {t_a} = \dfrac{u}{g}$
Now, again using Newton’s second law of equation, we get
$\Rightarrow {v^2} - {u^2} = 2gh$
As $v$ = 0 then $u = \sqrt {2gh}$
Then the time of ascent is ${t_a} = \dfrac{{\sqrt {2gh} }}{g}$………………………. (1)
Now, during descent the body will not oppose the gravity and undergo acceleration. Then using Newton’s equation of motion, we get
${v^2} = {u^2} + 2gh$
Now, here in this case the final velocity of ascent will become the initial velocity of descent therefore, $u$ = 0 on substituting in above equation, we get
$\Rightarrow {v^2} = 2gh$
$\Rightarrow v = \sqrt {2gh}$ …………………………… (2)
Now, using Newton’s first equation of motion and substituting $u$ = 0, we get
$v = u + g{t_d}$, td is the time of descent.
$\Rightarrow v = g{t_d}$
So, time of descent is ${t_d} = \dfrac{v}{g}$
Put the value of $v$ from equation (2) in above equation, we get
${t_d} = \dfrac{{\sqrt {2gh} }}{g}$ ………………… (3)
From equation (2) and (3), we get
$\Rightarrow {t_d} = {t_a}$
Hence,option (A) is correct.

Note
It should be noticed that during ascent the body will move in the opposite direction to the gravity due to which it undergoes retardation so we use negative sign of $g$ in newton’s equation of motion. But during the descent the body moves in the same direction as that of gravity so it undergoes acceleration and we use the positive sign of $g$ in Newton’s equation of motion in this case.