AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Velocity of projection,
v=2√gh
∴‌‌vx=v‌cos‌θ=2√gh‌cos‌θ
∴ Time taken by the projectile to cover interwall distance,
t=‌

2h

vx

=‌

2h

2√gh‌cos‌θ

⇒‌‌t=√‌

h

g

⋅sec‌θ...(i)
Vertical velocity at the top of the wall is given as
vy′2=vy2−2gh=(√2gh‌sin‌θ)2−2gh
=4ghsin2θ−2gh
vy22=2gh(2sin2θ−1)
∴‌‌vy′=√2gh(2sin2θ−1)
∴‌‌t=‌

2vy′

g

t=‌

2√2gh(2sin2θ−1)

g

...(ii)
From Eqs. (i) and (ii), we get
√‌

h

g

‌sec‌θ=‌

2√2gh(2sin2θ−1)

g

Squaring both side,
⇒‌

h

g

sec2θ=‌

4×2gh(2sin2θ−1)

g2

⇒1=8cos2θ(2sin2θ−1)
⇒1=8cos2θ[2(1−cos2θ)−1]
⇒16cos4θ−8cos2θ+1=0
(4cos2θ−1)2=0
⇒‌‌cos2θ=‌

1

4

⇒‌‌cos‌θ=‌

1

2

=cos‌60∘⇒θ=60∘
∴ From Eq. (i),
t=√‌

h

g

‌sec‌60∘=√‌

h

g

⋅2=√‌

4h

g