Asked by Sam

An archer shoots an arrow at a 78.0 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.)

Answers

Answered by Henry
Range = Vo^2*sin(2A)/g = 78 m.
34^2*sin(2A)/9.8 = 78
Solve for A.
Answered by Anonymous
449
Answered by kennny
20.695
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