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=‌