An archer shoots an arrow at a 72.0 m distant target, the bull's-eye of which is at same height as the release height of the arrow.

Question

(a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 36.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)
Please help me !

Answer 1

you know the range is given by

R = v^2/g sin2θ

so, plugging in your numbers,

36^2/9.8 sin2θ = 72