Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 10. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?

Answer 1

Check the picture below.

so, the maximum height is reached at the parabola's vertex, where x = how many seconds it took, y = how hight it went.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{lccclll} h = &{{ -16}}t^2&{{ +36}}t&{{ +10}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

$\bf \left(-\cfrac{36^2}{2(-16)}~~,~~10-\cfrac{36^2}{4(-16)} \right)\implies \left( \cfrac{1296}{32}~~,~~10+\cfrac{81}{4} \right) \\\\\\ \left( \cfrac{81}{2}~~,~~\cfrac{121}{4} \right)\implies \left(\stackrel{\textit{how many seconds}}{40\frac{1}{2}}~~,~~\stackrel{\textit{how many feet up}}{30\frac{1}{4}} \right)$