The position of a projectile launched from the origin at t = 0 is given by r → = 4 0 i ⁡ ^ + 5 0 j ⁡ ^ m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms -2 ).

Step-by-step solution

From question, Horizontal velocity (initial), u x = 40 2 = 20   m / s Vertical velocity (initial), 50 = u y t + 1 2 g t 2 ⇒ 50 =   u y × 2 + 1 2 ( - 10 ) × 4 50 = 2 u y - 20 u y = 70 2 = 35   m / s ∴    tan θ = u y u x = 35 20 = 7 4 ⇒ θ = tan - 1 7 4