Question

An archer shoots an arrow at an 82.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow.

(a)
At what angle in degrees must the arrow be released to hit the bull's-eye if its initial speed is 40.0 m/s?

Answer 1

Step-by-step explanation:

We can use the following kinematic equation to solve this problem:

y = y0 + tanθ(x - x0) - (gx²)/(2v₀²cos²θ)

where

y = 0 (since the target is at the same height as the release height)

y0 = 0

x0 = 0

x = 82.0 m

v₀ = 40.0 m/s

g = 9.81 m/s²

We want to solve for θ.

Rearranging the equation and substituting the values, we get:

tanθ = (xg)/(2v₀²)

θ = tan⁻¹[(xg)/(2v₀²)]

θ = tan⁻¹[(82.0 m)(9.81 m/s²)/(2(40.0 m/s)²)]

θ ≈ 18.1°

Therefore, the archer must release the arrow at an angle of approximately 18.1 degrees to hit the bull's-eye.