In the following problems the air resistance is constant (air resistance always opposes the motion).
(a) A ball is thrown vertically upward. If time of ascent is `t_(1)` and time of descent is `t_(2)`, which time is greater ?
(b) A ball is thrown vertically upward with speed `u` and on returning to the ground, its speed is `v_(0)`. Which speed is greater ?
(c) Two balls `A` and `B (m_(A) gt m_(B))` are thrown vertically upward with same speed. Which ball `A` or `B` (heavier or lighter) wil attain greater height ?

Explanation

(a) If you ask this question , majority of students will say that time of ascent `(t_(1))` is greater than time of descent `(t_(2))` , but this is incorrect. In fact time of descent is greater than time of ascent `(t_(2) gt t_(1))`. This can be proved in following manner.
Let the air resistance br `R` (constant in magnitude) and same for all bodies (given in problem).
Upward motion : `h = (1)/(2) a' t_(1)^(2)` (i)
Downward motion : `h = (1)/(2) at_(2)^(2)` (ii)
`(1)/(2) a' t_(1)^(2) = (1)/(2) at_(2)^(2)`
`(t_(2)^(2))/(t_(1)^(2)) = (a')/(a) = (g + (R)/(m))/( g - (R)/(m))`
` (t_(2))/(t_(1)) = sqrt((g + (R)/(m))/( g - (R)/(m)))`
Clearly , `t_(2) gt t_(1)`.
Time of descent `(t_(2))` gt Time of ascent `(t_(1))`
Think of two vehicles that are moving with same velocity. To stop the vehicles , on one vehicle `4 ` persons ( of equal strength) are employed and on another vehicle `7` persons are employed . The vehicle on which `7` persons are working will be stopped quickly . in the same manner , when the ball is moving up , forces opposing the motion are weight of hte ball and air resistance , while in downward motion opposing force is air resistance , obviously `t_(1) lt t_(2)`.
(b) Upward motion :
`v^(2) = u^(2) - 2 a' h` (i)
`0 = u^(2) - 2 a' h`
Downward motion :
`v^(2) = u^(2) + 2 ah`
`v_(0)^(2) = 0 + 2 ah` (ii)
`(v_(0)^(2))/(u^(2)) = (a)/(a')`
`(v_(0))/(u) = sqrt((g - (R)/(m))/(g + (R)/(m)))`
`v_(0) lt u` , speed `u` is greater.
( c) Ball A : retardation `a'_(A) = g + (R)/(m_(A))`
Ball B : retardation `a'_(B) = g + (R)/(m_(B))`
`v^(2) = u^(2) - 2a'h`
Ball A : `0 = u^(2) - 2 a'_(A) h_(A)` (i)
Ball B : `0 = u^(2) - 2 a'_(B) h_(B)` (ii)
` 2 a'_(A) h_(A) = 2 a'_(B) h_(B)`
`(h_(A))/(h_(B)) = (a'_(B))/(a'_(A)) = (g + (R)/(m_(B)))/(g + (R)/(m_(A)))`
`m_(A) gt m_(B)`
`(R)/(m_(A)) lt (R)/(m_(B))`
`(g + (R)/(m_(A))) lt ( g + (R)/(m_(B)))`
`h_(A) gt h_(B)`
The heavier ball (A) will attain greater height.
OR
The retardation of heavier ball will be smaller , hence it will attain greater height.