Check the picture below.
since the cliff is 48 feet tall, thus that its initial height.
[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&8\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&48\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+8t+48[/tex]
well, as you can see in the picture, its maximum is at its vertex, so
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+8}x\stackrel{\stackrel{c}{\downarrow }}{+48} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 8}{2(-16)}~~~~ ,~~~~ 48-\cfrac{ (8)^2}{4(-16)}\right)\implies \left(\cfrac{1}{4}~~,~~48-\cfrac{64}{-64} \right)[/tex]
[tex]\left( \frac{1}{4}~~,~~48+1 \right)\implies \stackrel{\textit{the ball is at highest}}{\left(\frac{1}{4}~~,~~\stackrel{\downarrow }{49} \right)}[/tex]
well, it hits the ground when y = 0.
[tex]\stackrel{h(t)}{0}=-16t^2+8t+48\implies 0=8(-2t^2+t+6)\implies 2t^2-t-6=0 \\\\\\ (2t+3)(t-2)=0\implies t= \begin{cases} -\frac{3}{2}\\\\ 2~~\textit{\large \checkmark} \end{cases}[/tex]
notice, we didn't use the negative value, since "t" must be greater than 0.
