A boy releases a toy plane from the top of a high hill of height H. Hill is so high that gravity varies with height from ground as `g = g_(0) (1 - (2h)/(R))`, where h is the height from ground and R is the radius of earth. The engine of toy plane accelerates it in horizontal direction with acceleration `a_(x) = bt^(2)`. Find the position from the foot of hill where the plane lands and the time after which it lands.

Answer

`[(bR^(2))/(48 g_(0)^(2)) [ ln ((R + sqrt(2RH - h^(2)))/(R - H))]^(2), sqrt((R)/(2g_(0)))ln ((R + sqrt(2RH - H^(2)))/(R - H))]`