Question

A rocket is fired upward from some initial distance above the ground. Its height is h, above the ground t seconds after it was fired, is given h=-16t^2+96t+3952. What was the rockets maximum height? How many seconds did it take to reach maximum height? After it is fired, how many seconds does it take to hit the ground?

Answer 1

So..hmm if you check the picture below, that's when it reaches the highest point

$\bf \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} h=&-16t^2&+96t&+3952\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so the rocket's maximum height was $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$  and it took $\bf -\cfrac{{{ b}}}{2{{ a}}}$ minutes to get there

now, when does it hit the ground? well, when f(x) = 0, that is, $\bf 0=-16t^2+96t+3952$

solve for "t"