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When a cannonball is fired, the equation of its pathway can be modeled by h = -16t² + 128t. Find the maximum height of the cannonball

User Igagis
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2 Answers

1 vote
h = 4ft per second it carries

User Carrein
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3 votes
so hmmm check the picture below

the leading term's coefficient for that quadratic is negative, meaning is opening downwards

so.. hmm where's the vertex anyway? well


\bf \textit{vertex of a parabola}\\ \quad \\ h=-16t^2+128t\\\\\\ \begin{array}{lccclll} h=&-16t^2&+128t&+0\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

so,the cannonball reaches a maximum height of
\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}
When a cannonball is fired, the equation of its pathway can be modeled by h = -16t-example-1
User Hanaan Rosenthal
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