Question

An archer shoots an arrow at a 67.0 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 36.0 m/s? (b) There is a large tree halfway between the archer and the target with an overhanging branch 4.79 m above the release height of the arrow. Will the arrow go over or under the branch? O over under

Answer 1

a)

range, R = u^2 sin 2x / g

67 = 36^2 sin (2x) /9.8

x = 15.22 deg

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b)max height is givenby

H = ( u sin x) ^2 / 2g

H = (36 sin 15.22)^2/19.6

H = 4.557 m

no the archer can't made the given distnace