Question

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of $2m/{s^2}$. The ratio of time of ascent to time of descent is ($g = 10m/{s^2}$ )
$
  A){\text{ }}1:1 \\
  B)\sqrt {\dfrac{2}{3}} \\
  C){\text{ }}\dfrac{2}{3} \\
  D){\text{ }}\sqrt {\dfrac{3}{2}} \\
    \\
 $

Answer 1

Hint: Here, we use the Newton’s equation of motion,
Expression be $s = ut + \dfrac{1}{2}a{t^2}$
Here, few replacements changes to h, and a changes to g because there is a gravitational force.
Expression be $h = ut + \dfrac{1}{2}g{t^2}$
The sign of acceleration of gravity changes with the condition
Given: acceleration due to gravity for upwards motion is $10m/{s^2} + 2m/{s^2} = 12m/{s^2}$ acceleration due to gravity for downwards motion is $10m/{s^2} - 2m/{s^2} = 8m/{s^2}$

Complete step by step solution:
Final velocity, v=$0m/s$ , for upward time represented by t
For upward motion, we have to find the value of initial velocity so we use this expression
$v = u + at$
$0 = u - gt$ (here, acceleration changes to –g because of the against the gravitational force)
Here, $u = gt$
Now, we find the distance,
${v^2} - {u^2} = 2gh$
$ \Rightarrow {(0)^2} - {u^2} = 2(12)h $
$ \Rightarrow {u^2} = 24h $
$ \Rightarrow h = \dfrac{{{u^2}}}{{24}} $
Above we find the value of initial velocity, now put the value of u in upward equation
$ h = \dfrac{{{{\left( {gt} \right)}^2}}}{{24}} = \dfrac{{{{\left( {12t} \right)}^2}}}{{24}} $
 $ \Rightarrow h = \dfrac{{144{t^2}}}{{24}} $
$ \Rightarrow h = 6{t^2} $
Here, we get the value of h also,
Now, for downward motion
For downward motion, Initial velocity, u= $0m/s$
Here time, represented in downward condition is T
Now, we calculate the value of h
$ h = ut + \dfrac{1}{2}g{t^2} $
$ \Rightarrow h = 0(t) + \dfrac{1}{2} \times 8 \times {T^2} $
$ \Rightarrow h = 4{T^2} $
Now, we equate the value of h in upward direction and downward direction
In downward direction the time represented by T , in upward direction the time represented by t.
Now, equating the value of h
\[h = 6{t^2}({\text{for upward motion) = 4}}{{\text{T}}^2}{\text{ (for downward motion)}}\]
Now,
$
  6{t^2} = 4{T^2} \\
 $
Now, the ratio be ascent to descent means upward to downward
 $ \dfrac{{{t^2}}}{{{T^2}}} = \dfrac{4}{6} $
  $ \Rightarrow \dfrac{{{t^2}}}{{{T^2}}} = \dfrac{2}{3} $
 $ \Rightarrow \dfrac{t}{T} = \sqrt {\dfrac{2}{3}} $
Here, option B is correct.

Note: The time of ascent is equal to the time of descent as air resistance acts in both cases.
That is, the time of the upward direction is equal to the time of the upward direction as air resistance acts.