TS EAMCET 18-July-2022 Shift 2 Solved Paper

The given situation is shown below.
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+‌

1

2

axt2
⇒‌‌R=vt+0‌‌(∵ax=0)
⇒‌‌R=vt
⇒R=√gH√‌

H

g

[From Eqs. (i) and (ii)]
⇒R=√‌

gH2

g

=H
⇒R=H