Question

A body is projected vertically up with certain velocity. At a point 'P' in its path, the ratio of its potential to kinetic energies is 9:16. The ratio of velocity of projection to velocity at 'P' is,

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Answer 1

• Kinetic Energy , K.E = $\sf \dfrac{1}{2}mv^2$
• Potential Energy , P.E = mgH

At point " P " ,

$:\implies \sf \dfrac{P.E}{K.E}=\dfrac{9}{16}$

$:\implies \sf \dfrac{mgH}{\frac{1}{2}mv_p^2}=\dfrac{9}{16}$

$:\implies \sf mgH=\dfrac{9}{32}mv_p^2$

Kinetic energy at point " P " = $\sf \dfrac{1}{2}mv_p^2$

$:\implies \sf \dfrac{1}{2}mv^2-mgH=\dfrac{1}{2}mv_p^2$

$:\implies \sf \dfrac{1}{2}mv^2=mgH+\dfrac{1}{2}mv_p^2$

$:\implies \sf \dfrac{1}{2}mv^2=\dfrac{9}{32}mv_p^2+\dfrac{1}{2}mv_p^2$

$:\implies \sf \dfrac{1}{2}mv^2=\dfrac{25}{32}mv_p^2$

$:\implies \sf \dfrac{v^2}{v_P^2}=\dfrac{25}{16}$

$:\implies \sf \pink{\dfrac{v}{v_p}=\dfrac{5}{4}}\ \; \bigstar$