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Mathematics 7 Online
OpenStudy (anonymous):

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. What is the maximum height of the particle? 128 feet 224 feet 272 feet 484 feet

OpenStudy (anonymous):

@iGreen

OpenStudy (igreen):

Okay, it's the same as the last one. First, we make the equation. \(\sf H(t) = -16t^2 + vt + s\) Where 'v' is the velocity(speed) and 's' is the starting height.

OpenStudy (igreen):

So plug in 160 for 's' and 144 for 'v'..what do you get?

OpenStudy (anonymous):

\[H(t)=-16t ^{2}+144(t)+160\]

OpenStudy (anonymous):

how do we figure out t

OpenStudy (igreen):

Correct! Now plug it into the equation \(\sf\dfrac{-b}{2a}\) to find the x-value of the vertex.

OpenStudy (igreen):

\(\sf H(t) = \color{red}{-16}t^2 + \color{lime}{144}t + \color{blue}{160}\) \(\sf \color{red} ax^2 + \color{lime} bx + \color{blue} c\)

OpenStudy (igreen):

So what are 'a' and 'b'?

OpenStudy (anonymous):

\[\frac{ -144 }{ 2(-16) }\]

OpenStudy (anonymous):

4.5 right?

OpenStudy (anonymous):

-4.5

OpenStudy (igreen):

Yep, now plug that back into the original equation for 't'.

OpenStudy (igreen):

\(\sf H(t) = -16t^2 + 144t + 160\)

OpenStudy (igreen):

Tell me what you get.

OpenStudy (igreen):

Oh wait, nevermind..4.5 was correct..lol.

OpenStudy (anonymous):

oh ok one sec

OpenStudy (anonymous):

\[H(t)=-16(4.5)^{2}+144(4.5)+160\]

OpenStudy (anonymous):

@iGreen sry my connection stopped

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