Let in 2 s body reaches upto point A and after one more sec upto point B. Total time of ascent for a body is given 3 s
i.e., t=
usinθ
g
=3 ∴ usinθ=10×3 usinθ=30 .....(i) Horizontal component of velocity remains always constant ucosθ=vcos30°....(ii) For vertical upward motion between point O and A vsin30°=usinθ−g×2[Usingv=u−gt] vsin30°=30−20[ As usinθ=30] ∴ v=20m/s Substituting this value v, we get in Eq. (ii) ucosθ=20cos30°=10√3 .....(iii) From Eqs. (i) and (iii) u=20√3 and θ=60°