A particle is projected from the surface of earth of mass M and radius R with speed `v`. Suppose it travels a distance `x(lt lt R)` when its speed becomes `v` to `v//2` and `y(lt lt R)` when speed changes from `v//2` to 0. Similarly, the corresponding times are suppose `t_(1) and t_(2)`. Then
`{:(,"Column-I",,"Column-II"),("(A)",x//y,"(p)",=1),("(B)",t_(1)//t_(2),"(r)",gt 1),(,,"(r)",lt 1):}`

Answer

`(A rarr q,B rarrp)`

Explanation

Given, `g=(GM)/(R^(2))="constant" " "(becausex lt ltR and Y ltltR)`
When speed become `v` to `(v)/(2)`
`((v)/(2))=v-gxxt_(1)rArrt_(1)=(v)/(2g)`
and `((v)/(2))^(2)=v^(2)-2gxrArr(3v^(2))/(8g)`
Similarly, `0=(v)/(2)-gxxt_(2)rArrt_(2)=(v)/(2g)`
and `0=((v)/(2))^(2)-2gyrArr y=(v^(2))/(8g)`
Clearly `t_(1) = t_(2), x gt y`.