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JEE PYQ: Motion in a Straight Line - Question ID 2add5106d1ca (JEE Main 2021)

ID: 2add5106d1caJEE Main 2021Single Correct MCQ

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching

{h \over 3}

in both the directions.

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Step-by-step Explanation

u = \sqrt {2gh}

Now,

S = {h \over 3}

a =

-

S = ut + {1 \over 2}a{t^2}

{h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}

{t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0

From quadratic equation

{t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}

{{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}

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