The angle of projection of the projectile launched is 52.4⁰
Explanation:
Let the initial velocity be x₀
According to the question, x₀cosθ = 5m/s
And x₀sinθ = 6.5m/s
Angle of projection = ?
Let the angle of projection be θ
So,
x₀sinθ / x₀cosθ = [tex]\frac{6.5}{5}[/tex]
We know,
sinθ / cosθ = tanθ
Thus,
tanθ = [tex]\frac{6.5}{5}[/tex]
tanθ = 1.3
θ = tan⁻¹ (1.3)
θ = 52.4⁰
Thus, angle of projection of the projectile launched is 52.4⁰