Question

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the object is 150 feet.
After how many seconds does the object reach its maximum height?

Answer 1

namely what is "t" when h(t) is 150, so

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+96t+6\implies \stackrel{h(t)}{150}=-16t^2+96t+6\implies 16t^2-96t+144=0 \\\\\\ t^2-6t+9=0\implies (t-3)(t-3)=0\implies \boxed{t=3}$

Answer 2

Answer:

3 seconds

Step-by-step explanation: