Question

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

s(t) (in feet)
of the ball from the ground is given by
s(t) = 132 + 48t − 16t2.
Find the maximum height attained by the ball.

Answer 1

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})`

Answer 2

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.