Kinematics

JEE Main level, single correct, medium

170 of 308

Question 170 of 308

An object of mass 5kg5kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10N10N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g=10ms−2g=10{ms}^{-2}].

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Answer: (B) 2:3\sqrt{2}:\sqrt{3}

Net force while going upward will be mg+R=50+10=60Nmg+R=50+10=60N
⇒aup=605=12ms−2\Rightarrow {a}_{up}=\dfrac{60}{5}=12m{s}^{-2}
Similarly, Net force while going downward will be mg−R=40Nmg-R=40N
⇒ad=405=8ms−2\Rightarrow {a}_{d}=\dfrac{40}{5}=8m{s}^{-2}
Since
h=12at2⇒auptup2adtd2=1⇒tuptd=812=23h=\dfrac{1}{2}a{t}^{2} \Rightarrow \dfrac{{a}_{up}{{t}_{up}}^{2}}{{a}_{d}{{t}_{d}}^{2}}=1\Rightarrow \dfrac{{t}_{up}}{{t}_{d}}=\sqrt{\dfrac{8}{12}}=\sqrt{\dfrac{2}{3}}

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