An archer shoots an arrow at a 49.0 m distant target, the bull\'s-eye of the tar

Question

An archer shoots an arrow at a 49.0 m distant target, the bull's-eye of the target is at same height as the release height of the arrow. At what angle must the arrow be released to hit the bull's-eye if its initial speed is 71.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. Submit Answer Tries 0/8 There is a large tree halfway between the archer and the target with an overhanging horizontal branch with the bottom of the branch at 0.76 m above the release height of the arrow, and the diameter of the branch is 0.063 m. Will the arrow go over or under or hit the branch? Neglect the thickness of the arrow. Go over the branch O Hit the branch Go under the branch Submit Answer Tries 0/1 This discussion is closed Send Feedback

Explanation / Answer

a) 1/2 arcsin (Rg/Vi^2) = 1/2 arcsin ((49 x 9.8)/71^2) = 2.73 degrees above horizontal.

b) Initial vertical V component = (sin 2.73 x 71) = 3.39 m/s

Max. height attained = (v^2/2g) = (3.39^2/19.6) = 0.585 m above release, < 0.76m

Go under the branch.

Formula for (a) derived from the range formula, Range = (Vi^2(sin 2* angle))/g.

g = 9.8 used.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.