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When a body is projected vertically up from the ground it's potential energy and kinetic energy at point P are in ratio 1:5.If same body is projected with

half the previous velocity , then at the same point P , ratio of its potential energy and kinetic energy would be
do you know the answers, i jzt want to confirm my solution
yes its 2:1
is ur ans right???
plzzzzz tell


Energy at point P = K.E. + P.E. 
let H be the maximum point where the body will reach and h be the point P
so K.E. at P = mg(H - h)
⇒ P.E./K.E. = mgh /mg(H-h) = 1/5
⇒h / H-h = 1/5 ⇒ 5h = H-h ⇒ H = 5h +h 
so H = 6h
let u be 0 
so v²-u²=2gH
= v²=2gH
⇒ H = v²/2g
but it is given that velocity is halved so 
H = (v/2)² / 2g
    = v²/4/2g
    v²=2gh + u² = 2gh ( since u=0)
   so H = 2gh/4/2g = 2gh/4 x 1/2g = h/4
so ration of P.E and K.E. = h/ (h/4 - h) = h/ (6h/4 - h) 
since H = 6h
so , ratio = h/(6h/4 - h) = 2/1 or 2:1
1 5 1
thank u
but its tough
even my teachers cant solve it
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