Question 1150174
Write a model that describes the trajectory of the rock. 
<pre>
{{{h(t)=h(0)+v(0)t-16t^2}}}

{{{h(t)=80+64t-16t^2}}}

</pre>How many seconds until the max height is reached?<pre> 

The easiest way is to use the vertex formula, for the maximum height 
is reached at the vertex which has t coordinate.  We put the 
quadratic in standard form at²+bt+c

{{{h(t)=-16t^2+64t+80}}}

{{{-b/(2a)=(-(64))/(2(-16))=(-64)/(-32) = 2}}}{{{seconds)

Another way:
The maximum height will be reached in half the time it takes the
rock to get back down even with Jacob, who is 80 feet above ground
at the bottom of the cliff.  So we set h(t)=80 and solve for t,
the take half.

{{{80=80+64t-16t^2}}}
{{{0=64t-16t^2}}}
{{{0=16t(4-t)}}}
{{{16t=0}}}  {{{4-t=0}}}
{{{t=0}}}    {{{4=t}}}

It take 4 seconds for the rock to get back even with Jacob, so it
reached its maximum height in half that time or 2 seconds.  But
the vertex formula is the easiest.

</pre>What will be the maximum height?<pre> 

The rock's maximum height is reached when t=2 seconds.

{{{h(t)=80+64t-16t^2}}}
{{{h(2)=80+64(2)-16(2)^2}}}
{{{h(2)=80+128-16(4)}}}
{{{h(2)=80+128-64}}}
{{{h(2)=144}}}{{{ft}}}

</pre>How many seconds until the rock hits the ground?<pre>

That's when h(t) = 0

We set h(t)=0 and solve for t

{{{h(t)=80+64t-16t^2}}}
{{{0=80+64t-16t^2}}}

Divide through by 16

{{{0=5+4t-t^2}}}
{{{0=(5-t)(1+t)}}}
{{{0=5-t}}}    {{{0=1+t}}}
{{{t=5}}}      {{{-1=t}}}

Ignore the negative answer.

5 seconds.

</pre>How high is the rock after 3 seconds?<pre>

This is h(3)

{{{h(t)=80+64t-16t^2}}}
{{{h(3)=80+64(3)-16(3)^2}}}
{{{h(3)=80+192-16(9)}}}
{{{h(3)=80+192-144}}}
{{{h(3)=128}}}{{{ft}}}

Edwin</pre>