what is a? An archer shoots an arrow at a 77.0 m distant target, the bull\'s-eye

Question

what is a? An archer shoots an arrow at a 77.0 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 34.0 m/s? (Although neglected here. the atmosphere provides significant lift to real arrows.) X degree (b) There is a large tree hallway between tile archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the anew go over or under the branch? Over under

Explanation / Answer

Given that

horizontal range is R = u^2*sin(2*theta)/g

here u = 34 m/s

g = 9.8 m/s^2

R = 77 m

77 = 34^2*sin(2*theta)/9.8

sin(2*theta) = 77*9.8/(34*34) = 0.652

2*theta = asin(0.652) = 40.75

theta = 20.37 degrees

B) Hmax = u^2*sin^2(20.37)/(2*g)= 34*34*sin^2(20.37)/(2*9.8) = 7.14 m

7.14>3.5 m

hence the answer is over

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