A particle is thrown upards from ground. It experiences a constant resistance force which can produce retardation of 2m/s 2. The ratio of time of ascent to the time of descent is (take g=10m/s 2 ).

							
Hi there.
In First Case, When the Ball is thrown upwards.
Initial Velocity = u
Final velocity(v) = 0
[∵ The ball will stop before coming down to the surface]
Height at which the ball reaches before coming down = S m
Time taken by the Particles to reach the height (or Time of Ascent) = t s.
Acceleration = -g
[∵ The ball is thrown against the gravity] 
From the Question, it experiences the air Resistance of 2 m/s².
∴ Acceleration of the Particles = -g - 2
 = - (g + 2) m/s².
Now, Using the Equation of the motion,
 v = u + at
∴ 0 + u + [-(g + 2)]t
∴ u = t(g + 2)
⇒ t = u/(g + 2)
Hence, time taken by the Particle to reach the certain height (or Time of Ascent) is u/(g + 2) seconds.
Now, Again using the Equation of the Motion,
v² - u² = 2aS
⇒ (0)² - (u)² = 2[-(g + 2)S]
⇒ u² = 2S(g + 2)
⇒ S = u²/(g + 2)

Now, In Second Case, (When the ball comes down),
Initial Velocity(u) = 0
Final Velocity = v
Time taken by the Particles to falls (or Time of Descent) = T s.
Height or  Distance covered(S) = u²/(g + 2) 
Acceleration of the Particle = g m/s²
∴ The Particles experienced the Air Resistance, 
 ∴ Acceleration = (g - 2) m/s².

∵ S = ut + 1/2 × aT²
∴ u²/(g + 2) = 0 × T + 1/2 × (g - 2) × T²
⇒ T = 
Hence, the Time taken by the Particles to falls down (or Time of Descent) is 

Now, A/c to the Question,We have to calculate the ratio of the time of ascent to the time of descent.
∴ Time of Ascent/Time of Descent = t/T 
    = u/(g + 2) ÷ u/√(g² - 4)
    = 
    = 
    = 
    = 
    = 
    = √96/12
    = 4√6/12
    = √6/3
∴ t : T = √6 : 3

Hence, the ratio of the time of the Ascent to the time of descent is √6 : 3
Note ⇒ Time of Ascent is always equals to the time of descent only when there is no air resistance thus no loss of energy. But since there is the loss of energy by the air resistance, it cannot be possible.