# How long it takes the arrow to come back to the ground if Andrew is an avid archer and he launches an arrow that takes a parabolic path, modeled by the equation y=-4.9t^2+48t?

Nov 28, 2014

Since $y = 0$ at the ground level, we have

$0 = - 4.9 {t}^{2} + 48 t = t \left(- 4.9 t + 48\right)$

$\implies \left\{\begin{matrix}t = 0 \\ t = \frac{48}{4.9} \approx 9.8\end{matrix}\right.$

Hence, it takes approximately $9.8$ s.

I hope that this was helpful.