An archer shoots an arrow at a 76.0 m distant target, the bull\'s-eye of which i

Question

An archer shoots an arrow at a 76.0 m distant target, the bull's-eye of which 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 37.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)

(b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Explanation / Answer

Part A)

Apply R = vo^2sin(2angle)/g

76 = (37)^2(sin 2(angle))/(9.8

angle = 16.5 degrees

Part B)

Apply H = vo^2sin^2(angle)/2g

H = (37^2)(sin^2(16.5))/(2)(9.8)

H = 5.63 m

H is greater than 3.5, so the arrow goes over the branch

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