Question

A body is thrown vertically upward in air and air resistance is taken into account. The time of ascent is ${t_1}$ and time of descent is ${t_2}$, then which of the following is true
(A) ${t_1} = {t_2}$
(B) ${t_1} > {t_2}$
(C) ${t_1} < {t_2}$
(D) Cannot be predicted

Answer 1

Hint:In order to solve this question, we should know that air resistance force always acts in a direction opposite to that of direction of motion of the body, so here we will find effective acceleration of body in both cases and then using newton’s equation of motion to find relation between time of ascent and time of descent.

Formula used:
$v = u + at$
${v^2} - {u^2} = 2aS$
$S = ut + \dfrac{1}{2}a{t^2}$
where,
v is the final velocity, u is the initial velocity of the body
S is the distance covered, a is the acceleration of the body and t denotes time taken by the body.


Complete answer:
Let us assume the acceleration due to air resistance force is ‘a’ and acceleration due to gravity is ‘g’ and total height reached by the body is H, during ascent and descent of the particle the net acceleration acting on the particle is shown in the diagram as

Now, During ascent time taken by the body to rise with initial velocity u and net acceleration ${a_{net}} = g + a$ is
using equation $v = u + at$ as final velocity during ascent is zero and net acceleration will be taken as negative so ${t_1} = \dfrac{u}{{g + a}}$ and height H reached by the body can be calculated using ${v^2} - {u^2} = 2aS$ we get
$
  {u^2} = 2(g + a)H \\
   \Rightarrow H = \dfrac{{{u^2}}}{{2(g + a)}} \to (i) \\
 $
Now, during the descent the distance covered by the body will be H with acceleration $a{'_{net}} = g - a$ with time ${t_2}$ so using newton’s equation of motion as
$S = ut + \dfrac{1}{2}a{t^2}$ and since initial velocity is zero during descent so we get,
$H = \dfrac{1}{2}(g - a){t_2}^2$ using the value of H from equation (i) we get,
$
  \dfrac{{{u^2}}}{{2(g + a)}} = \dfrac{1}{2}(g - a){t_2}^2 \\
  {t_2} = \dfrac{u}{{g + a}}\sqrt {\dfrac{{g + a}}{{g - a}}} \\
 $
using the value ${t_1} = \dfrac{u}{{g + a}}$ we get,
${t_2} = {t_1}\sqrt {\dfrac{{g + a}}{{g - a}}} $
now since, we know that $\dfrac{{g + a}}{{g - a}} > 1$ so,
${t_2} > {t_1}$ or ${t_1} < {t_2}$
So, the time of descent will be larger than the time of ascent.
Hence, the correct answer is ${t_1} > {t_2}$

Hence, the correct option is Option (B).


Note:Always pay attention while calculating the acceleration values during ascent and descent. Also consider the effect of gravity on the motion of the body during ascent and descent.