Answer to Question #218083 in Mechanics | Relativity for Busisiweleticia.bl

Question #218083
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. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 37m/s?
1
Expert's answer
2021-07-19T15:42:39-0400

X- direction

Newton's second law

Sx=ut+12at2a=0Sx=ut75=37(cosθ)tS_x=ut+\frac{1}{2}at^2\\a=0 \\S_x=ut\\75=37(cos\theta)t

Y-direction


Sy=uyt+12at2S_y=u_yt+\frac{1}{2}at^2


Sy=0,a=g,uy=usinθ=37sinθS_y=0,a=-g,u_y=usin\theta=37sin\theta

0=37(sinθ)t+12(9.8)t20=37(sin\theta)t+\frac{1}{2}(-9.8)t^2

0=(37sinθ)+12(9.8)tt=709.81sinθ0=(37sin\theta)+\frac{1}{2}(-9.8)t\\t=709.81sin\theta\\

75=37cosθ×(709.81×sinθ)75=24509.8sinθcosθ75=37cos\theta \times(709.81\times sin\theta)\\75=24509.8sin\theta cos\theta\\

sin2θ=2sinθcosθsin2\theta=2sin\theta cos\theta

75=13131.4851sin2θ75=13131.4851sin2\theta

θ=16.5°\theta=16.5°


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