The position of a projectile launched from the origin at t=0 is given by r=(40i+50j) m at t=2 s. If the projectile was launched at an angle θ from the horizontal, then θ is (Take g=10 ms⁻²).

According to the question, At t=2 sec
Horizontal distance covered, l = ucosθt = 2ucosθ
Vertical distance covered, h = usinθt - (1/2)gt²
Since, l=40 and h=50
We solve these 2 equations simultaneously for u and θ.
We get, tanθ = 5/4
Or, θ = tan⁻¹(5/4)