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A body is lauched from the earth's surfa...

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 (a) the height to which the body will rise. (ii) the radius of curvature of trajectory at its top point.

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The correct Answer is:
`(a) h=(sqrt(7))/2+1)R,(b)1.13 R`
(a) Applying conservation of angular momentum between point `P` and `Q`

`m(v_(0)cos30^(@))R=mvr`
`v=(sqrt(3))/2(v_(0)R)/r...(1)`
Applying conservation of mechanical energy.
`-(GMm)/R+1/2mv_(0)^(2)=-(GMm)/r+1/2mv^(2)...(2)`
From `(1)` and `(2)`
`-(GMm)/R+1/2mv_(0)^(2)=-(GMm)/r+1/2m(3v_(0)^(2)R^(2))/(4r^(2))...(3)`
given that `v_(0)=sqrt(3/2(GM)/R)...(4)`
From `(3)` and `(4)`
`4r^(2)-16Rr+9R^(2)=0`
`r=2R+(sqrt(7))/2RimpliesR+h=2R+(sqrt(7))/2R`
`h=(sqrt(7)/2+1)Rimpliesh=sqrt(7)/2R`
(b) Radius of curvature `=(v^(2))/g=((sqrt(3)v_(0)R)/(2r^(2)))^(2)//(GM)/(R^(2))`
radius of curvature=`3/4(v_(0)^(2)R^(2))/(GM)`
`=3/4xx((3/2sqrt(GM)/R))^(2)xx(R^(2))/(GM)=9/8R=1.125 R`
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