A projectile is launched from the ground with an initial horizontal velocity of 5.0m/s [right] and an initial vertical velocity of 6.5m/s [up]. At what angle from vertical is it launched?

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 =

We know,

sinθ / cosθ = tanθ

Thus,

tanθ =

tanθ = 1.3

θ = tan⁻¹ (1.3)

θ = 52.4⁰

Thus, angle of projection of the projectile launched is 52.4⁰