A projectile of mass `m` is fired from the surface of the earth at an angle `alpha = 60^(@)` from the vertical. The initial speed `upsilon_(0)` is equal to `sqrt((GM_(e))/(R_(e))`. How high does the projectile rise ? Neglect air resistance and the earth's rotation.

Options

A. `(R_(e))/2`
B. `(R_(e))/5`
C. `(R_(e))/4`
D. `(R_(e))/8`

Answer

A

Explanation

Let `v` be the speed of the projectile athighest point and `r_(max)` its distance from the centre of the earth. Applying conservation of angular momentum and mechanical energy,
`mv_(0)sinalpha=mvr_(max)........(i)`
`1/2mv_(0)^(2)-(GM_(e)m)/(R_(e))=1/2mv^(2)-(GM_(e)m)/(r_(max)).....(2)`
Solving these two equation with given data we get, `r_(max)=(3R_(e))/2`
or the maximum height `h_(max)=r_(max)-R_(e)=(R_(e))/2`