Question

A rocket is launched vertically from the ground with an initial velocity of 64 ft/sec.

Answer 1

we know the rocket was launched from the ground, meaning its initial height is 0, thus

$~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&64\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&0\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t$

Check the picture below.