Originally Posted by **Soroban**

Hello, kkmarie9009!

The function is a quadratic (second degree).

. . Its graph is a down-opening parabola.

. . The maximum height occurs at the vertex of the parabola.

The formula for the vertex of a parabola is: $\displaystyle x = \frac{-b}{2a}$

Our parabola has: $\displaystyle a = -4.9,\;b = 24,\;c = 2$

The vertex is at: $\displaystyle t \:= \:\frac{-24}{2(-4.9)} \:=\:\frac{120}{49} \:= \:2.4489795 $

(a) Therefore, the rock reaches maximum height at $\displaystyle t = 2.45$ seconds.

If $\displaystyle t = \frac{120}{49}$, then: $\displaystyle h\;=\;2 + 24\left(\frac{120}{49}\right) - 4.9\left(\frac{120}{49}\right)^2 \;=\;31.3877551$

(b) Therefore, the maximum height is: $\displaystyle h \:= \:31.39$ m.

If the rock hits the ground, its height is zero.

. . Solve the quadratic: $\displaystyle 2 + 24t - 4.9t^2\:=\:0\quad\Rightarrow\quad 4.9t^2 - 24t - 2\:=\:0$

Quadratic Formula: $\displaystyle t \;= \;\frac{-(-24) \pm \sqrt{(-24)^2 - 4(4.9)(-2)}}{2(4.9)}$

Hence: $\displaystyle t\;= \;\frac{24 \pm \sqrt{615.2}}{9.8} \;= \;4.979920979$

(c) Therefore, the rock hits the ground in about $\displaystyle 4.98$ seconds.