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Mathematics 15 Online
OpenStudy (anonymous):

You drop a rock off a bridge. The rock's height, h (in feet above the water), after t seconds is modeled by h=-16 t^2+541. What is the height of the rock after 2 seconds?

OpenStudy (compassionate):

\(\Huge{\color{purple}{\textbf{W}} \color{orange}{\cal{E}} \color{green}{\mathbb{L}} \color{blue}{\mathsf{C}} \color{maroon}{\rm{O}} \color{red}{\tt{M}} \color{gold}{\tt{E}} \space \color{orchid}{\mathbf{T}} \color{Navy}{\mathsf{O}} \space \color{OrangeRed}{\boldsymbol{O}} \color{Olive}{\mathbf{P}} \color{Lime}{\textbf{E}} \color{DarkOrchid}{\mathsf{N}} \color{Tan}{\mathtt{S}} \color{magenta}{\mathbb{T}} \color{goldenrod}{\mathsf{U}} \color{ForestGreen}{\textbf{D}} \color{Salmon}{\mathsf{Y}} \ddot \smile }\) h=-16 t^2+541 There are two methods. You can use the quadratic equation to locate h, or factor it. Lets try factoring it. Looking at it you see you can't. So, just use the vertex formula. b/-2a

OpenStudy (jdoe0001):

well, I guess you may not need ... that one after all anyhow

OpenStudy (anonymous):

sorry messed that up. So if we drop a rock of a cliff with have initial height but no initial velocity. so Displacement=s or the height after two seconds

OpenStudy (jdoe0001):

"t" in the equation stands for Seconds, how how high it's after 2 seconds? well, set t = 2 and solve for "h"

OpenStudy (anonymous):

so with the equation your velocity initial is zero and your initial height is the last term so plug in t and you'll get the height of the rock

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