Question

A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. How much time does it take for the rock to hit the ground?

Answer 1

Check the picture below.

so it hits the ground when y = 0, thus

$\bf ~~~~~~\textit{initial velocity}\\\\ \begin{array}{llll} ~~~~~~\textit{in feet}\\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{16}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{5}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases}$

$\bf h(t)=-16t^2+16t+5\implies \stackrel{h(t)}{0}=-16t^2+16t+5 \\\\\\ 16t^2-16t-5=0\implies (4t+1)(4t-5)=0\\\\ -------------------------------\\\\ 4t+1=0\implies 4t=-1\implies t=-\cfrac{1}{4}\\\\ -------------------------------\\\\ 4t-5=0\implies 4t=5\implies t=\cfrac{5}{4}\implies \boxed{t=1\frac{1}{4}}$

since "t" is seconds it took, it can't be a negative amount.