Question

A particle is thrown upwards from ground.it experiences a constant air resistance which can produce retardation of 2m/s^2

Answer 1

When a particle in thrown vetically upward and it faces air resistance causing retardation of 2m/s^2
Then, its upward or vertical motion will be,
During its upward motion, it will face a total retardation of (9.8 + 2) m/s^2
And during its downward motion, it will face a acceleration of (9.8 - 2) m/s^2.

Answer 2

your complete question is ------>A particle is thrown upwards from the ground. it experiances a constant resistance force which can produce retardation 2m/sec sqr. the ratio of the time of ascent to the time of the descent is__? [g=10m/sec sqr] ?

solution :- for upward motion ,
acceleration = - (g + 2) m/s² [ negative sign shows Acceleration is acting downward direction.]
so, use formula, v = u + at, here t is time taken for ascending motion
at heighest position , v = 0
0 = u - (g + 2)t
u = (g + 2)t
t = u/(g + 2) ------(1)
now, v² = u² + 2aS
0 = u² - 2(g + 2)S
S = u²/2(g + 2) -------(2)

For downward motion,
acceleration = (g - 2) m/s²
initial velocity ,u = 0
so, S = uT + 1/2 aT² , here T is time taken for descending motion
from equation (2),
u²/2(g + 2) = 0 + 1/2 (g - 2)T²
From equation (1),
(g + 2)²t²/2(g + 2) = 1/2(g -2)T²
t²/T² = (g - 2)/(g + 2)
square root both sides,
t/T = √(g - 2)/(g + 2)
t/T = √(10 - 2)/(10 + 2) [ assume g = 10 m/s²
t/T = √{8/12} = √2/√3
Hence, ratio of time taken in ascent and descent is √2 : √3