When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then, x1 : x2 will be

The motion of the object shot in two cases can be depicted as belowUsing third equation of motion,  v2 = u2 - 2gh      ......(i) As the object stops finally, so v = 0For inclined motion, g = gsinθ and h = xSubstituting these values in Eq. (i), we get⇒  u2=2gsinθx⇒  x=u22gsinθFor case I  x1=u22gsin600For case II  x2=u22gsin300∵  x1x2=u22gsin60°×2gsin30°u2=12×23=13 or 1:3