A particle is thrown upwards from ground .it experiences a constant air resistance which can produce retardation of 2m/s2 opposite to the direction of velocity of particle . The ratio of time of ascent and descent is ??

Let A be the retardation force. Time for ascent be TTime for descent be T`Now, we know that u = 0 So we have two equations:- S = 1/2 (g-a) T² (for ascent) and S = 1/2 (g+a) T`² (for descent)Divide them and we will get –T/T` = sqrt of (g-a) / sqrt of (g+a) T/T` = sqrt of (10-2) / sqrt of (10+2) Therefore, ratio of T and T` is -Sqrt of 2 / sqrt of 3 .