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?

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.[tex]\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)[/tex][tex]\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)[/tex]