Question

A ball is thrown from a height of 160 feet with an initial downward velocity of 12/fts. The ball's height h (in feet) after t seconds is given by the following: h = 160 − 12t - 16t^2
How long after the ball is thrown does it hit the ground?

Answer 1

Check the picture below, so it hits the ground when h = 0, therefore,

$\bf \stackrel{h}{0}=160-12t-16t^2\implies 16t^2+12t-160=0\\\\\\ 4t^2+3t-40=0 \\\\\\ \textit{now let's use the quadratic formula} \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \begin{array}{lcccl} y=& 4 t^2& +3 t& -40\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf t=\cfrac{-3\pm\sqrt{3^2-4(4)(-40)}}{2(4)}\implies t=\cfrac{-3\pm\sqrt{9+640}}{8} \\\\\\ t= \begin{cases} \frac{-3+\sqrt{649}}{8}\\\\ \frac{-3-\sqrt{649}}{8} \end{cases}\approx \begin{cases} \boxed{2.81}\\\\ -3.56 \end{cases}$