now, if the ball is just free-falling, is simply dropped from 100ft, then the initial velocity is 0, and the initial height is 100, and it hits the ground when its height reaches 0.
[tex]\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array}
\quad
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}
\end{cases}
\\\\\\
\begin{cases}
h_o=100\\
v_o=0\\
h(t)=0
\end{cases}\implies 0=-16t^2+0t+100\implies 16t^2=100[/tex]
[tex]\bf t^2=\cfrac{100}{16}\implies t^2=\cfrac{25}{4}\implies t=\sqrt{\cfrac{25}{4}}\implies t=\cfrac{\sqrt{25}}{\sqrt{4}}\implies t=\cfrac{5}{2}~secs[/tex]
