A particle is projected from the surface of earth with speed v. Suppose it travel a distance x when its speed become v to `(v)/(2)` and y when speed changes form `(v)/(2)` to 0. Similarly, the corresponding times are suppose `t_(1) " and "t_(2)`. Then

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`.