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padilas [110]
3 years ago
7

A ball is thrown straight up from the top of a building 132 ft. tall with an initial velocity of 48 ft per second. The distance

Mathematics
2 answers:
arsen [322]3 years ago
6 0

The line of symmetry through the parabola described by s(t) is

... t = -48/(2×(-16)) = 1.5

This is the time at which the ball reaches its maximum height.

... s(1.5) = 132 + 48×1.5 -16×1.5²

... s(1.5) = 132 + 72 - 36 = 168


The maximum height attained by the ball is 168 ft.

egoroff_w [7]3 years ago
6 0

check the picture below.


so the maximum height of the ball is obtained at the vertex's y-coordinate, hmm what is it anyway?


\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
s(t)=132+48t-16t^2
\\\\\\
s(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+48}t\stackrel{\stackrel{c}{\downarrow }}{+132}
\qquad \qquad 
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~  c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( \qquad ,~~132-\cfrac{48^2}{4(-16)} \right)\implies \left( \qquad ,~~132+\cfrac{2304}{64} \right)
\\\\\\
\left( \qquad ,~~132+36 \right)\implies (\quad ,~\stackrel{feet}{168})

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What is the answer to this?<br> Thank you
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