JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Kinematics

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An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use

g = 10

^{- 2}

]

(A) $1 : 1$
(B) $2 : 3$ ✓ Correct
(C) $3 : 2$
(D) $2 : 3$

Answer: (B)

Solution

During ascent, net downward force =

m g + f = 50 + 10 = 60

N, so

a_{1} = 12

m/s². During descent, net downward force =

m g - f = 50 - 10 = 40

N, so

a_{2} = 8

m/s². Using

h = \frac{1}{2} a t^{2}

for same height:

a_{1} t_{1}^{2} = a_{2} t_{2}^{2}

, giving

\frac{t _{1}}{t _{2}} = \frac{a _{2}}{a _{1}} = \frac{8}{12} = \frac{2}{3} = \frac{2}{3} = 2 : 3

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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