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. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

a) Θ = 18.5°

b) h = 6.26 m, 3.50 m, arrow goes over the branch

Explanation:

having the following dа ta:

Vo = 35 m/s

Θ = ?

horizontal distance = 75 m

a)

using the following equations:

Voy = Vo * (sin Θ) = 35 * (sin Θ)

Vox = Vo * (cos Θ) = 35 * (cos Θ)

horizontal distance to target = 75 = Vox * (2t); where t = Voy/g

replacing values:

75 = Vox * (2/g) * (Voy) = 2 * (Vox) * (Voy) / 9.8 = (Vo) ²*[2 * (sin Θ) * (cos Θ) ]/g = (Vo) ² (sin2Θ) / g

solving and using trigonometric identities:

sin2Θ = 75 * (g) / (Vo) ² = 75 * (9.8) / (35) ² = 0.6

2Θ = 36.91°

Θ = 18.5°

b)

The time to reach the maximum height will be equal to:

t = Voy/g = 35 * (sin18.5°) / 9.8 = 1.13 s

and the maximum height will be equal to:

h = 1/2gt² = (0.5) * (9.8) * (1.13) ² = 6.26 m, 3.50 m, arrow goes over the branch