Respuesta :
[tex]\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{116}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{101}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+116t+101\implies \stackrel{h(t)}{0}=-16t^2+116t+101[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+116}x\stackrel{\stackrel{c}{\downarrow }}{+101} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-116\pm\sqrt{116^2-4(-16)(101)}}{2(-16)} \\\\\\ t=\cfrac{-116\pm\sqrt{116^2-4(-16)(101)}}{2(-16)}\implies t=\cfrac{-116\pm\sqrt{116^2+6464}}{-32} \\\\\\ t\approx \begin{cases} -0.7855696910943375\\ \boxed{8.0355696910943375} \end{cases}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+116}x\stackrel{\stackrel{c}{\downarrow }}{+101} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t=\cfrac{-116\pm\sqrt{116^2-4(-16)(101)}}{2(-16)} \\\\\\ t=\cfrac{-116\pm\sqrt{116^2-4(-16)(101)}}{2(-16)}\implies t=\cfrac{-116\pm\sqrt{116^2+6464}}{-32} \\\\\\ t\approx \begin{cases} -0.7855696910943375\\ \boxed{8.0355696910943375} \end{cases}[/tex]
It take the rocket 8 seconds to hit the ground after it is launched.
Let h represent the height of the rocket at time t.
The height (h) is given by:
h = –16t² + vt + c
Since the initial velocity (v) = 116 ft/s, the initial height c = 101 ft, hence:
h = –16t² + 116t + 101
Hence, the rocket touches the ground at time h = 0, hence:
0 = -16t² + 116t + 101
Hence, t = 8 seconds.
Therefore it take the rocket 8 seconds to hit the ground after it is launched.
Find out more at: https://brainly.com/question/20689870
