Question

A body is projected vertically upwards with some velocity. At a point on its path 'p', the ratio of P.E. to K.Ē. is 9:16. Then the ratio of the velocity of projection to the velocity at that point 'p' is
1) 3:4
2)5:4
3) 9:25
4) 25:16
​

Answer 1

Answer :

2) 5 : 4

Explanation :

At point ' P ' ,

$\to \sf \dfrac{P.E}{K.E}=\dfrac{9}{16}\\\\\to \sf \dfrac{mgH}{\frac{1}{2}mv_p^2}=\dfrac{9}{16}\\\\\pink{\leadsto \sf mgH=\dfrac{9}{32}mv_p^2}\ \; \bigstar$

Kinetic energy at point ' P ' is given by ,

$\to \sf \dfrac{1}{2}mv_p^2=\dfrac{1}{2}mv^2-mgH\\\\\to \sf \dfrac{1}{2}mv_p^2=\dfrac{1}{2}mv^2-\dfrac{9}{32}mv_p^2\\\\\to \sf \dfrac{1}{2}mv^2=\dfrac{1}{2}mv_p^2+\dfrac{9}{32}mv_p^2\\\\\to \sf v^2=v_p^2+\dfrac{9}{16}v_p^2\\\\\to \sf v^2=v_p^2\left(1+\dfrac{9}{16}\right)\\\\\to \sf \dfrac{v^2}{v_p^2}=\dfrac{25}{16}\\\\\to \sf \dfrac{v}{v_p}=\sqrt{\dfrac{25}{16}}\\\\ \sf \orange{\leadsto \dfrac{v}{v_p}=\dfrac{5}{4}}\ \; \bigstar$

Answer 2

Answer :

2) 5 : 4

Explanation :

At point ' P ' ,

Kinetic energy at point ' P ' is given by ,