Question 112941
The basic formula to use is {{{s(t)=(1/2)gt^2+v[o]t+s[o]}}} where s(t) is the distance or height function with respect to time, {{{s[o]}}} is the initial or starting height, {{{v[o]}}} is the initial velocity, {{{g}}} is the constant acceleration due to gravity, and {{{t}}} is the time.


The first thing to realize is that {{{g=-32(ft/sec^2)}}}.  Why is it negative?  Because the gravity vector is downward and you are throwing the rock upwards.


The other constants are given:


{{{v[o]=64ft/sec}}}
{{{s[o]=25ft}}}.  You could make an argument that the initial height is 25 feet plus the distance from the roof of the building to the end of your outstretched arm at the moment you release the rock, but I think for the purposes of this problem you can ignore all of that.


The first part of the problem asks how high the rock will be after 1 second.  Simply substitute 1 for t and evaluate:


{{{s(t)=-16t^2+64t+25}}}
{{{s(1)=-16(1)^2+64(1)+25}}}
{{{s(1)=-16+64+25=73}}}

So after 1 second, the rock will be at 73 feet above the ground.


Second part of the problem:


The graph of {{{s(t)}}} is a parabola that is convex down (the coefficient on the {{{x^2}}} term is < 0).  If we can find the t value of the vertex, we will have the time when the rock reaches maximum height.


The vertex of {{{y=ax^2+bx+c}}} is given by {{{-b/2a}}}.  So for our problem, that is {{{-64/2(-16)=-64/-32=2}}}.  So maximum height will be reached in 2 seconds.


Third part:

The maximum height is given by evaluating the function at the time we just determined in the previous part of the problem:


{{{s(t)=-16t^2+64t+25}}}
{{{s(2)=-16(2)^2+64(2)+25}}}
{{{s(1)=-64+128+25=89}}}


So the maximum height is 89 feet.


Hope that helps.
John