Three particles are projected vertically upward from a point on the surface of earth with velocities
`v_(1)=sqrt((2gR)/(3)),v_(2)sqrt(gR),v_(3)sqrt((4gR)/(3))`
respectively, where g is acceleation due to gravity on the surface of earth. If the maximum height attained are `h_(1),h_(2) " and " h_(3)` respectively, then `h_(1):h_(2):h_(3)` is

Options

A. `1:2:3`
B. `2:3:4`
C. `1:2:4`
D. `1:3:5`

Answer

C

Explanation

( c) Loss of K.E.= Gain of P.E.
`(1)/(2)mv^(2)=(mgh)/(1+(h)/( R))`
`(1)/(2)mv_(1)^(2)=(mgh_(1))/(1+(h_(1))/( R))`
`(1)/(2)mv_(1)^(2)=(mgRh_(1))/(R+h_(1))`
`(1)/(2)m(2gR)/(3)=(mgRh_(1))/(R+h_(1))`....(i)
`(1)/(2)mgR=(mgRh_(2))/(R+h_(2))`....(ii)
`(1)/(2)m(4gR)/(3)=(mgRh_(3))/(R+h_(3))`....(iii)
` or h_(1)=( r)/(2)` (From equation (i))
`h_(2)=R` (From equation (ii))
`h_(3)=2R` (From equation (iii))
`h_(1):h_(2):h_(3)=( R)/(2):R:2R=(1)/(2):1:2=1:2:4`