Question

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?

Answer 1

The answer to your question is angle = 18.46°

Step-by-step explanation:

Data

d = 75 m

v₀ = 35 m/s

α = ?

Formula

`d = (vo^(2)sin2\alpha)/(g)`

solve for sin2α

sin 2α =
`(dg)/(vo^(2))`

Substitution

sin 2α =
`(75(9.81))/(35^(2))`

Simplify

sin 2α =
`(735.75)/(1225)`

Divide

sin 2α = 0.600

Get sin⁻¹

2α = 36.9°

Divide by 2

α = 18.5°