Question

An arrow leaves a bow at an angle of O in reference to the horizontal. The initial velocity is vo = 111 feet per second. The arrow hits a target 327 feet away. Find onevalue of 0 if the range of the arrow is given by the equation r = 32voʻsin(20). Express your answer in degrees or radians rounded to the nearest hundredth.=

Answer 1

The range of the projectile is the displacement in the horizontal direction. It is given by the equation;

`\begin{gathered} r=(1)/(32)v^2_o\sin (2\theta) \\ \text{Where;} \\ v_o=\text{initial velocity} \end{gathered}`

Thus, given;

`\begin{gathered} r=327ft \\ v_o=111(ft)/(s) \end{gathered}`

We have;

`\begin{gathered} 327=(1)/(32)(111)^2\sin (2\theta) \\ 327=(1)/(32)(12321)\sin (2\theta) \\ 327=385.03\sin (2\theta) \\ \sin (2\theta)=(327)/(385.03) \\ \sin (2\theta)=0.8493 \\ 2\theta=\sin ^(-1)(0.8493) \end{gathered}`
`\begin{gathered} 2\theta=58.1336 \\ \theta=(58.1336)/(2) \\ \theta=29.07^o \end{gathered}`