H=7 + 29t-16t^2 find all values of t for which the - id7291624866817, orlando19882000

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Brendah4962

17.01.2022

$\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}
\end{cases}\\\\
-------------------------------\\\\$

$\bf h(t)=7+29t-16t^2\qquad h(t)=19\implies 19=7+29t-16t^2
\\\\\\
16t^2-29t+12=0\impliedby \textit{now, let's use the quadratic formula}
\\\\\\
t=\cfrac{-(-29)\pm\sqrt{(-29)^2-4(16)(12)}}{2(16)}\implies t=\cfrac{29\pm\sqrt{841-768}}{32}
\\\\\\
t=\cfrac{29\pm\sqrt{73}}{32}\implies t\approx 
\begin{cases}
1.17\\
0.64
\end{cases}$

so hmm check the picture below, thus the ball hits 19 feet of height twice, once on the way up, and once on the way down, at about 0.64 seconds and at 1.17 seconds.
H=7 + 29t-16t^2 find all values of t for which the balls height is 19ft

H=7 + 29t-16t^2 find all values of t for which the balls height is 19ft

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H=7 + 29t-16t^2 find all values of t for which the balls height is 19ft

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