Question

If a ball is thrown upward at 64 feet per second from a height

of 4 feet, the height of the ball can be modeled by
S = 4 +64t - 16t² feet, where t is the number of seconds
after the ball is thrown. How long after the ball is thrown is the
height 32 feet?

Answer 1

`S=4+64t-16t^2\implies \stackrel{\textit{at a height of 32}}{32=4+64t-16t^2}\implies 16t^2-64t-4+32=0 \\\\\\ 16t^2-64t+28=0\implies 4(4t^2-16t+7)=0\implies 4t^2-16t+7=0 \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{4}t^2\stackrel{\stackrel{b}{\downarrow }}{-16}t\stackrel{\stackrel{c}{\downarrow }}{+7}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`t= \cfrac{ - (-16) \pm \sqrt { (-16)^2 -4(4)(7)}}{2(4)} \implies t = \cfrac{ 16 \pm \sqrt { 256 -112}}{ 8 } \\\\\\ t= \cfrac{ 16 \pm \sqrt { 144 }}{ 8 }\implies t=\cfrac{16\pm 12}{8}\implies t= \begin{cases} (1)/(2)&\textit{on the way up}\\\\ (7)/(2)&\textit{on the way down} \end{cases}`