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.
Solution
$u = \sqrt {2gh}$<br><br>Now, <br><br>$S = {h \over 3}$<br><br>a = $-$g<br><br>$S = ut + {1 \over 2}a{t^2}$<br><br>${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$<br><br>${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$<br><br>From quadratic equation<br><br>$${t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$$<br><br>$${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$
About this question
Subject: Physics · Chapter: Kinematics · Topic: Motion in a Straight Line
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