Question

if a projectile is launched with an intial velocity of 13 meters per second, what angle is required to achieve a range of 15 meters?

Answer 1

Answer:

30° or 60°

Step-by-step explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

Given R = 15 m and v₀ = 13 m/s:

15 m = (13 m/s)² sin(2θ) / (9.8 m/s²)

sin(2θ) = 0.870

2θ = 60.4° or 119.6°

θ = 30.2° or 59.8°

Rounded, the launch angle required is either 30° or 60°.