A rock is launched straight up into the air from the edge of a \(100 foot\) cliff with an initial vertical velocity of \(15 \frac {ft}{sec}\). The rock rises to its peak and then falls past the cliff to the river at the bottom of the cliff.

How far above the cliff did the rock climb?

The initial positive is \(100 ft\).
The initial vertical velocity is \(15 \frac {ft}{sec}\)

That is gives the following function for vertical position above the river.

\[ h(t)= -\frac {9.81}{2} t^2 + 15 t + 100 \]

This is a quadratic function. The formula has a negative leading coefficient, so we know there is a maximum value occuring at \(t = \frac {-15}{2 \cdot (-\frac {9.81}{2})} \approx 1.529 sec\).

\[ h(1.529) \approx -\frac {9.81}{2} \cdot (1.529)^2 + 15 \cdot 1.529 + 100 \approx 111.468 ft \]
When does the rock hit the water?

We want the time when the height is \(0 ft\).

\[ h(t)= -\frac {9.81}{2} t^2 + 15 t + 100 = 0 \]
\[ -\frac {9.81}{2} t^2 + 15 t + 100 = 0 \]

This is a quadratic equation with a negative leading coefficient and positive constant term, which means there are two real solutions.

\[ t = \frac {-15 \pm \sqrt {15^2 - 4 \cdot (-4.905) \cdot 100}}{-9.81} \approx \frac {-15 \pm \sqrt {2187}}{-9.81} \]
\[ t \approx 6.296164302 \, \text { or } \, t \approx -3.238060327 \]

Since out stopwatch is moving forward from \(0\), the rock hits the watter at approximately \(6.296 sec\)

2025-01-07 02:39:05