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

Question

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?

Answer 1

X- direction

Newton's second law

$S_x=ut+\frac{1}{2}at^2\\a=0 \\S_x=ut\\75=37(cos\theta)t$

Y-direction

$S_y=u_yt+\frac{1}{2}at^2$

$$S_y=0,a=-g,u_y=usin\theta=37sin\theta$$

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

$0=(37sin\theta)+\frac{1}{2}(-9.8)t\\t=709.81sin\theta\\$

$75=37cos\theta \times(709.81\times sin\theta)\\75=24509.8sin\theta cos\theta\\$

$sin2\theta=2sin\theta cos\theta$

$75=13131.4851sin2\theta$

$\theta=16.5°$