Kinematics

JEE Main level, single correct, medium

173 of 308

Question 173 of 308

A ball is thrown up with a certain velocity so that it reaches a height h.h. Find the ratio of the two different times of the ball reaching h3\dfrac{h}{3} in both the directions.

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Answer: (C) 3−23+2\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}

u=2ghu=\sqrt{2gh}
Now,
S=h3,a=−gS=\dfrac{h}{3},a=-g
S=ut+12at2S=ut+\dfrac{1}{2}a{t}^{2}
h3=2ght+12(−g)t2\dfrac{h}{3}=\sqrt{2gh}t+\dfrac{1}{2}\left(-g\right){t}^{2}
t2(g2)−2ght+h3=0{t}^{2}\left(\dfrac{g}{2}\right)-\sqrt{2gh}t+\dfrac{h}{3}=0
From quadratic equation
t1,t2=2gh±2gh−4g2h3g{t}_{1},{t}_{2}=\dfrac{\sqrt{2gh}\pm \sqrt{2gh-\dfrac{4g}{2}\dfrac{h}{3}}}{g}
t1t2=2gh−4gh32gh+4gh3\dfrac{{t}_{1}}{{t}_{2}}=\dfrac{\sqrt{2gh}-\sqrt{\dfrac{4gh}{3}}}{\sqrt{2gh}+\sqrt{\dfrac{4gh}{3}}}
=3−23+2=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}

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