Question

2. On top of an 64-foot mountain, a stone is thrown upward following a trajectory. The stone will fall into the lake after t seconds. The stone's height h, in feet above the surface of the lake is given by the equation h = -16t2 +64t + 64.

Answer 1

h = -16t² + 64t + 64

We want to know at what time the stone falls into the lake. If the stone is in the lake, its height is zero.

-16t² + 64t + 64 = 0

Applying the quadratic formula:

`\begin{gathered} t_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t_(1,2)=\frac{-64\pm\sqrt[]{64^2-4\cdot(-16)\cdot64}}{2\cdot(-16)} \\ t_(1,2)=\frac{-64\pm\sqrt[]{8192}}{-32} \\ t_1=(-64+90.5)/(-32)=-0.83 \\ t_2=(-64-90.5)/(-32)=4.83 \end{gathered}`

The negative result has no sense in the context of the problem. Then, it takes 4.83 seconds for the stone to fall into the lake