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.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?