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 initial velocity =u
Ascent : accleration =`g+2=12 m//s ^(2) `
Time of ascent `= (u)/(12) ` and max . Height `=h=(u^(2))/( 2 xx 12 ) =(u^(2))/(24) `
Descent : acceleration `= g-2=8m//s^(2) `
Time of descent =` sqrt (( 2h)/( 8)) =sqrt (( 2 xx (u^(2))/(24) )/(8)) =( u)/( 4sqrt6) " "," " "Ratio "=( "Time of asscent ")/("Time of descent ") =sqrt(( 6)/( 3)) =sqrt(( 2)/(3))`