Question

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?

Answer 1

`\bf ~~~~~~\textit{initial velocity}\\\\ \begin{array}{llll} ~~~~~~\textit{in feet}\\\\ h(t) = -16t^2+v_ot+h_o \\\\ \end{array} \quad \begin{cases} v_o=\stackrel{24}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{50}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at`

now, check the picture below, it reaches a maximum point at the vertex, where the seconds is the x-coordinate and the height is the y-coordinate.

so, let's use the coefficients to get the vertex out of that quadratic then,


`\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{lccclll} h(t) = &{{ -16}}t^2&{{ +24}}t&{{ +50}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)`


`\bf \left(-\cfrac{24}{2(-16)}~~,~~50-\cfrac{24^2}{4(-16)} \right)\implies \left( \cfrac{3}{4}~~,~~50- \cfrac{576}{-64}\right) \\\\\\ \left( \cfrac{3}{4}~~,~~50+ 9\right)\implies \left(\stackrel{\textit{seconds}}{(3)/(4)}~~,~~\stackrel{feet}{59} \right)`