A projectile rises upto a maximum height of `R//(1-K^(2))` where `K` is a constant and `R` is the radius of earth. If the velocity of projectile with which it should be fired upwards form the surface of the earth to reach this height is equal to the product of a coeffiecient and escape velocity then this coeffiecient is equal to:

Options

A. `K`
B. `K^(2)`
C. `K^(3//2)`
D. `K^(1//2)`

Answer

A

Explanation

`r=R/(1-K^(2))`.......(i)
`h=r-R=(K^(2)R)/((1-K^(2)))`........(ii)
`1/2mv^(2)=1/2m(K'v_(e))^(2)=(mgh)/(1+h//R)`
`=mgh(R/(R+h))=mgr(R/r)`
or, `1/2kK'^(2)xx2gR=mg(K^(2)R)/((1-K^(2))xx(1-K^(2)) [ :' v_(e)=sqrt(2gR)]`
From (i), (ii) `K'=K`