AP EAPCET 05-Jul-2022 Shift 1 Paper

According to law of conservation of energy at point B shown in the figure given in question part, Loss in PE= Gain in KE
‌⇒mg(H−h)=‌

1

2

mv2
‌⇒‌‌v=√2g(H−h)
Now, h=‌

1

2

gt2⇒t=√‌

˙2h

g

∴ Distance covered in horizontal portion,
s‌=v×t
‌=√2g(H−h)×√‌

2h

g

⇒‌‌s‌=√4h(H−h)......(i)
For maximum value of s
‌

ds

dh

=0
⇒‌

1

2√4h(H−h)

×4[(H−h)⋅1+h(−1)]=0
⇒2(H−2h)=0⇒H=2h
⇒h=‌

H

2

Substituting the value of h in Eq. (i), we get,
s‌=√4‌

H

2

(H−‌

H

2

)
‌=√2H⋅‌

H

2

=H