A projectile is fired over horizontal ground from the origin with an initial speed of 60 m/s. What firing angles will produce a range of 250 m?

Angle θ ≅ 21.5°

Explanation:

Given:

speed Vi = 60 m/s, Range R = 250 m, g=9.81 m/s²

To find:

Angle θ = ?

Sol: we know that Rang R = Vi² sin 2θ / g

⇒ Sin 2θ = g*R / Vi²

Sin 2θ = (9.81 m/s² * 250 m) / (60 m/s) ²

Sin 2θ = 0.68125

2θ = Sin ⁻¹ (0.68125)

2θ = 42.9413

θ = 42.9413 / 2

θ = 21.4706 ≅ 21.5°