Question

An arrow is shot vertically upward from a platform 20 ft high at a rate of 179ft / s * e When will the arrow hit the ground ? Use the formula h = - 16t ^ 2 + v_{0}*t + h_{0} Round your answer to the nearest tenth

Answer 1

The height attained by the arrow is modelled as,

$h(t)=-16t^2-v_{\circ}t+h_{\circ}$

Given that the initial height is 20 ft and the initial velocity is 179 ft per second,

$\begin{gathered} v_{\circ}=179 \\ h_{\circ}=20 \end{gathered}$

Substitute the values,

$\begin{gathered} h(t)=-16t^2-(179)t+(20) \\ h(t)=-16t^2-179t+20 \end{gathered}$

When the arrow hits the ground its height will become zero,

$\begin{gathered} h(t)=0 \\ -16t^2-179t+20=0 \end{gathered}$

Applying the quadratic formula,

$\begin{gathered} t=\frac{-(-179)\pm\sqrt[]{(-179)^2-4(-16)(20)}}{2(-16)} \\ t=\frac{179\pm\sqrt[]{33321}}{-32} \\ t=\frac{179\pm182.54}{-32} \\ t=\frac{179+182.54}{-32},t=\frac{179-182.54}{-32} \\ t=-11.298,t=0.1106 \end{gathered}$

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.