A body is lauched from the earth's surface a an angle `alpha=30^(@)` to the horizontal at a speed `v_(0)=sqrt((1.5 GM)/R)`. Neglecting air resistance and earth's rotation, find the height to which the body will rise.

Answer

(i). `h=((sqrt(7))/(2)+1)R`
(ii). `1.13R`

Explanation

`v_(1)(H+R)=(3)/(2)sqrt((GM)/(2R))Rimpliesv_(1)=((3R)/(R+H))sqrt((Gm)/(8R))`
`v_(0)^(2)-(GM)/(2R)=v^(2)-(2GM)/((R+H))`
`-(GM)/(2R)=(9R^(2))/((R+H)^(2))(GM)/(8R)-(2GM)/(R+h)R`
`-1=(9R^(2))/(4(R+H))-(4R)/(R+H)`
`-1=(9R^(2)-16R^(2)-16RH)/(4(R+H)^(2))`
`4R^(2)+4H^(2)+8RH+9R^(2)-16R^(2)-16RH=0`
`4H^(2)-8RH-3R^(2)=0`
`=(8R+-sqrt(64R^(2)+48R^(2)))/(8)`
`=R+-Rsqrt(R^(2)+(3)/(4)R^(2))`
`=R+-(R)/(2)sqrt(7)`