A particle thrown up vertically reaches its highest point in time `t_(1)` and returns to the ground in a further time `t_(2)`. The air resistance exerts a constant force on the particle opposite to its direction of motion.

Options

A. `t_(1)=t_(2)`
B. `t_(1)gtt_(2)`
C. `t_(1)lt t_(2)`
D. Information is not sufficient to decide the relation between `t_(1)` and `t_(2)`

Answer

c

Explanation

When the particle moves upwards then its weight as the well as the air resistance both acts downwards hence the retardation for the upward journey can be calculated as follows:
`Mg+F=Ma_(1)`
`rArr a_(1)=g+(F)/(M)`
When particle moves downward then weight acts downwards but air resistance acts upwards, hence the acceleration during the downward journey can be calculated as follows:
`Mg-F-Ma_(2)`
`rArr a_(2)=g-(F)/(M)`
If h is the maximum height attained by the particle then we can write the following:
`h=(1)/(2)a_(1)t_(1)^(2)=(1)/(2)a_(2)t_(2)^(2)`
`(t_(2))/(t_(1))=(sqrt(a_(1))/(a_(2)))`
From equation 1 and 2 we can see that `a_(1)gta_(2)` hence `t_(2)gtt_(1)`