s. When will it reach its maximum height? How far above the ground will it be?

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jdoe0001

Is anyone good at math? I need help solving this problem.

A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be?

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jdoe0001

So hmm check the picture below

$\bf \text{initial velocity}\\
h(t) = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\ \quad \\

\begin{cases}
v_o=\textit{initial velocity of the object}\to &24\\
h_o=\textit{initial height of the object}\to &50\\
h=\textit{height of the object at "t" seconds}
\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\
y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the ball will reach its maximum height at $\bf -\cfrac{{{ b}}}{2{{ a}}}$ seconds

and will be $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$ feet above the ground
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