An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow.

d = 75 mv = 37 m/s

At what angle must the arrow be released to hit the bull’s-eye if i its initial speed is 37m/s

Equation for projectile motion on level ground:

d = v^2*sin(2*theta)/g

Thus, sin(2*theta) = d*g/v^2

Thus, 2*theta = sin^-1(d*g/v^2)

Thus, theta = .5*sin^-1(d*g/v^2)

Plugging in: theta = .5*sin^-1(75*9.81/37^2) = 16.25 degrees

Please feel free to approach me directly with physics problems in the future. Thanks.

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