Given, Initial velocity of projectile, u=40m∕s Angle, θ=30∘ Time of flight T=
2usinθ
g
=
2×40×1
10×2
=45(∵g=10m∕s2) It means projectile is at maximum height at t=2s. At maximum height vertical component of velocity is zero. Velocity at t=2s=Vx=ucosθ=40cos30∘ =20√3ms−1.