time in air: distance/speed=72/36=2s
hf=hi-vsinTheta*t-4.9t^2
hf=hi and you know time t, solve for theta
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.
(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.)
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