Answered

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A ball is thrown into the air with an initial velocity of 80 ft/sec. The height of the ball after t seconds in the air is given by h(t) = -16t? + 80t + 96. When will the ball be at its
highest point? Show all work.



Sagot :

Check the picture below.

[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+80}t\stackrel{\stackrel{c}{\downarrow }}{+96} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]

[tex]\left(-\cfrac{ 80}{2(-16)}~~~~ ,~~~~ 96-\cfrac{ (80)^2}{4(-16)}\right) \implies \left( - \cfrac{ 80 }{ -32 }~~,~~96 - \cfrac{ 6400 }{ -64 } \right) \\\\\\ \left(\cfrac{5}{2}~~,~~96+100 \right)\implies \stackrel{highest~point}{\stackrel{\qquad ~~ \downarrow }{\left(2\frac{1}{2}~~,~~196 \right)}}[/tex]

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