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(B) Let the maximum height reached by the ball be $H$. At the maximum height,the final velocity is $0$. Using the equation of motion $v_f^2 = v_i^2 -

Vedclass · Accepted Answer

$ \frac{v}{\sqrt{2}}, \frac{E}{2} $

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(B) Let the maximum height reached by the ball be $H$. At the maximum height,the final velocity is $0$. Using the equation of motion $v_f^2 = v_i^2 - 2gH$,we get $0 = v^2 - 2gH$,which implies $H = \frac{v^2}{2g}$.At half the height,$h = \frac{H}{2} = \frac{v^2}{4g}$.Using the equation $v'^2 = v^2 - 2gh$,the velocity $v'$ at height $h$ is $v'^2 = v^2 - 2g(\frac{v^2}{4g}) = v^2 - \frac{v^2}{2} = \frac{v^2}{2}$.Thus,$v' = \frac{v}{\sqrt{2}}$.The kinetic energy $E'$ at this point is $E' = \frac{1}{2}mv'^2 = \frac{1}{2}m(\frac{v^2}{2}) = \frac{1}{2}(\frac{1}{2}mv^2) = \frac{E}{2}$.

$A$ ball is thrown vertically upwards with a velocity $v$ and an initial kinetic energy $E$. When it reaches half way to the top of its flight,what are its velocity and kinetic energy respectively?

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