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Answer :
Check the picture below.
how long will it take for it to hit the ground, or namely, what is "t" when h(t) is 0.
[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&100\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&0\\ \qquad \textit{of the object}\\ h=\textit{object's height}&h\\ \qquad \textit{at "t" seconds} \end{cases}[/tex]
[tex]h(t)=-16t^2+100t+0\implies h(t)=-16t^2+100t\implies \stackrel{h(t)}{0}=-16t^2+100t \\\\\\ 16t^2-100t=0\implies 4t(4t-25)=0\implies \begin{cases} 4t=0\\ t=0\\[-0.5em] \hrulefill\\ 4t-25=0\\ 4t=25\\ t=\cfrac{25}{4}\\[1em] t=6\frac{1}{4} \end{cases}[/tex]
so h(t) is 0, as you saw on the picture, at two instances, once it was on the ground, 0 seconds, or t=0, and then t = 6.25, or 6 seconds and some change later.
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