The height attained by the arrow is modelled as,
[tex]h(t)=-16t^2-v_{\circ}t+h_{\circ}[/tex]
Given that the initial height is 20 ft and the initial velocity is 179 ft per second,
[tex]\begin{gathered} v_{\circ}=179 \\ h_{\circ}=20 \end{gathered}[/tex]
Substitute the values,
[tex]\begin{gathered} h(t)=-16t^2-(179)t+(20) \\ h(t)=-16t^2-179t+20 \end{gathered}[/tex]
When the arrow hits the ground its height will become zero,
[tex]\begin{gathered} h(t)=0 \\ -16t^2-179t+20=0 \end{gathered}[/tex]
Applying the quadratic formula,
[tex]\begin{gathered} t=\frac{-(-179)\pm\sqrt[]{(-179)^2-4(-16)(20)}}{2(-16)} \\ t=\frac{179\pm\sqrt[]{33321}}{-32} \\ t=\frac{179\pm182.54}{-32} \\ t=\frac{179+182.54}{-32},t=\frac{179-182.54}{-32} \\ t=-11.298,t=0.1106 \end{gathered}[/tex]
Since time cannot be negative, we have to neglect the negative value.
Thus, it can be concluded that the arrow will hit the ground after 0.1106 seconds approximately.
