An archer wishes to shoot an arrow at a target at eye level a distance of 50.0m away. If the initial speed imparted to the arrow is 70.0m/s, what angle should the arrow make with the horizontal as it is being shot?

Physics

Given:

Hmax (distance) = 50.0m

v ₀ = 70.0m/s

Required:

what angle should the arrow make with the horizontal as it is being shot

Solution:

Hmax = v ₀²sin²θ / 2g

sin²θ = 2gHmax / v₀²

sin²θ = 2 (9.81 m/s²) (50m) / (70 m/s) ²

sin²θ = 0.200

θ = 26.56°