A small object slides down with initial velocity equal to zero from the top of a smooth hill of height H. The other end of the hill is horizontal and is at height ‌
H
2
as shown in the figure. The horizontal distance covered by the object from the end of the hill to the ground is
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+‌
1
2
ayt2 ⇒‌‌‌
H
2
=0+‌
1
2
gt2‌‌(∵uy=0‌ at ‌P) ⇒‌‌t=√‌
2H
2g
⇒t=√‌
H
g
‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) ‌ and ‌‌‌x=uxt+‌