Let the speed at point P=v By the conservation of energy, ∴−∆U=∆K where, −∆U= decrease in potential energy and ∆K= increase in kinetic energy of object. ⇒−(Uf−Ui)=Kf−Ki ⇒−(mg
H
2
−mgH)=
1
2
mv2−
1
2
mu2 ⇒mg
H
2
=
1
2
mv2−0(∵u=0) ⇒v=√gH... (i) Now, when the object reaches point P , let its speed be v in horizontal direction and R be the distance from the end of the hill to the ground and also the time taken by the object to travel from P to Q be t . ∴⋅y=uyt+