Mathematics: Aprojectile is launched straight up with an initial velocity of 120 ft/s. if acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft?

Aprojectile is launched straight up with an initial

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greysonelmore1212

11.06.2020

$\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}\\
----------\\
v_o=120\\
h_o=0\\
h(t)=200
\end{cases}$

$\bf h(t)=-16t^2+120t\implies 200=-16t^2+120t
\\\\\\
16t^2-120t+200=0\implies 2t^2-15t+25=0
\\\\\\
(2t-5)(t-5)=0\implies 
\begin{cases}
2t-5=0\implies &t=\frac{5}{2}\\\\
t-5=0\implies &t=5
\end{cases}$

it reaches 200ft twice, on its way up, then on its way down, and 5/2 secs first, then on its way down at 5secs

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Ответ:

ItsLeyArt

16.03.2021

Mathematics

It's actually C

Step-by-step explanation:

don't forget about the probability of 0 too

you have to add the two probability formulas

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Aprojectile is launched straight up with an initial velocity of 120 ft/s. if acceleration due to gravity is –16 ft/s2, after about how many seconds will the object reach a height of 200 ft?

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