An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. at what angle (in degrees) must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s?

The answer to your question is angle = 18.46°

Explanation:

Data

d = 75 m

v₀ = 35 m/s

α = ?

Formula

solve for sin2α

                  sin 2α =

Substitution

                 sin 2α =

Simplify

                sin 2α =

Divide

                sin 2α = 0.600

Get sin⁻¹

                      2α = 36.9°

Divide by 2

                        α = 18.5°                

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.