(a) A ball is thrown vertically upward with speed `10 m//s` and it returns to the ground with speed `8 m//s`. Find the maximum height attained by the ball.
(b) A ball is thrown vertically upward and if air resistance is half of weight of the ball, find the ratio of time of ascent and time of descent.
(C) Two balls `A` and `B (m_(A) = 2 m_(B) = 2 m)` are thrown vertically upward. If air resistance is `mg//2`, find ratio of maximum height attained by them.

Explanation

(a) Upward motion :
R : air resistance
Retardation `a' = (mg + R)/(m) = g + (R)/(m)`
`0 = (10)^(2) - 2a'h = 100 - 2(g + (R )/(m))h`
`rArr gh + (R )/(m)h = 50` (i)
Downward direction :
Acceleration `a = (mg - R)/(m) = g - (R )/(m)`
`(8)^(2) = 0 + 2 a h`
`64 = 2(g - (R )/(h))h`
` = gh - (R)/(m) h = 32` (ii)
Adding (i) and (ii) , we get
`2gh = 50 + 32 = 82`
`h = (82)/(2g) = (82)/(20) = 4.1 m`
(b) Upward motion : `R = (mg)/(2)`
Retardation `a' = (mg + R)/(m) = (mg + (mg)/(2))/(m) = (3g)/(2)`
`h = (1)/(2) a' t_(1)^(2) = (3g)/(4) t_(1)^(2)` (iii)
Downward motion:
Acceleration `a = (mg - R)/(m) = (mg - (mg)/(2))/(m) = (g)/(2)`
`h = (1)/(2) at_(2)^(2) = (g)/(4) t_(2)^(2)` (iv)
`(3g)/(4) t_(1)^(2) = (g)/(4) t_(2)^(2)`
`(t_(1))/(t_(2)) = (1)/(sqrt(3))`
(c ) Upward motion `R = (mg)/(2) , mA = 2m , m_(B) = m`
Retardation `a'_(A) = (m_(A) g + R)/(m_(A)) = g + (R)/(m_(A))`
`= g + (mg)/( 2 xx 2m) = (5g)/(4)`
`0 = u^(2) - 2a'_(A) h_(A)` (v)
Ball B : retardation `a'_(B) = g + (R)/(m_(B)) = g + (mg)/( 2m) = (3g)/(2)`
`0 = u^(2) - 2a'_(B) h_(B)` (vi)
From (v) and (vi) , we have
`(h_(A))/(h_(B)) = (a'_(B))/(a'_(A)) = ((3g)/(2))/((5g)/(4)) = (6)/(5)`