A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of `2 ms^-2.` The ratio of time of ascent to time of descent 13 (`g = 10 ms^-2`)

Options

A. `1 : 1`
B. `sqrt((2)/(3))`
C. `(2)/(3)`
D. `sqrt((3)/(2))`

Answer

B

Explanation

Let `a` be the retardation produced by resistive force. Also let `t_a` and `t_d` be the time ascent and descent respectively.
If the particle rises upto a height `h`
Then `h = (1)/(2)(g + a) t_a^2` and `h = (1)/(2) (g - a)t_d^2`
`:. (t_a)/(t_d) = sqrt((g -a)/(g + a)) = sqrt((10 - 2)/(10 + 2)) = sqrt((2)/(3))`.