Question 53230
If you need a calculus way of solving this, let me know and I will edit this to include calculus.
a) v0=32; s0=0; 
{{{s=-16t^2+v0t+s0}}}
{{{s=-16t^2+32t+0}}}
{{{s=-16t^2+32t}}}
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b){{{s=-16(1)^2+32(1)}}}
{{{s=-16+32}}}
s=16 ft.
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c) The position s=0 when the ball hits the ground.
{{{0=-16t^2+32t}}}
{{{0=-16t(t-2)}}}
-16t=0
-16t/-16=0/-16
t=0s.  This is the time we threw the ball.
t-2=0
t-2+2=0+2
t=2 s.  This is the time that the ball hit the ground.
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d)This is a parabolic function and it's maximum value occurs at its vertex.  There are a couple of ways to find the vertex.  If your taking caluculus, this is where you would do something different than what I'm showing you now.
The x value of the vertex can be found by putting your function in standard form.
{{{s=at^2+bt+c}}}
{{{t=-b/(2*a)}}}
In your problem:
{{{t=-16t^2+32t}}}
a=-16, b=32
{{{t=-(32)/(2*-16)}}}
{{{t=(-32)/(-32)}}}
t=1 s.  (At 1 s. the ball is as high as it will go.)
Substitute that into your function to find out how high it will go.
{{{s=-16(1)^2+32(1)}}}
{{{s=-16(1)+32(1)}}}
s=-16+32
s=16 ft.  That is how high the ball will go in a perfect physics universe.