Question
A body is thrown upwords. If air resistance causing deceleration of $$5\,m/{s^2},$$ then ratio of time of ascent to time of descent is [take $$g = 10\,m/{s^2}$$ ]
A.
$$\sqrt {\frac{1}{2}} $$
B.
$$\sqrt {\frac{1}{{2.5}}} $$
C.
$$\sqrt {\frac{1}{3}} $$
D.
$$\sqrt {\frac{1}{5}} $$
Answer :
$$\sqrt {\frac{1}{3}} $$
Solution :
$$\eqalign{
& \frac{{{\text{Time of}}\,{\text{ascent}}}}{{{\text{Time of descent}}}} = \frac{{\left( {\frac{u}{{g + a}}} \right)}}{{\frac{u}{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }}}} \cr
& = \frac{{\sqrt {\left( {g + a} \right)\left( {g - a} \right)} }}{{\left( {g + a} \right)}} \cr
& = \sqrt {\frac{{g - a}}{{g + a}}} \cr
& = \sqrt {\frac{{10 - 5}}{{10 + 5}}} \cr
& = \sqrt {\frac{5}{{15}}} \cr
& = \sqrt {\frac{1}{3}} \cr} $$