Question

A missile is launched at an angle of `60^(@)` to the vertical with a velocity `sqrt(0.75 gR)` from the surface of the earth (R is the radius of the earth). Find its maximum height from the surface of earth. (Neglect air resistance and rotation of earth).

Answer 1

From conservation of mechanical energy
`(1)/(2)mv_(0)^(2)-(GMm)/(R )=(1)/(2)mv_(1)^(2)-(GMm)/(R+h)` ….(i)
Also from conservation of angular momentum
`mv_(0)R sin 60^(@) = mv_(1) (R+h)` ….(ii)
Solving (i) and (ii) and putting `v_(0) = sqrt((3GM)/(4R))`, we get
`h ~~ 0.25 R`.