Question 58644
Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 feet per second.  A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s=-16t2+v0t+s0

* 16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2)
* v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
* s0 is the initial distance above ground (in feet).  If you are standing on the ground, then s0=0

a) What is the function that describes this problem?
so=0 and vo=32
s=-16t^2+32t+0
{{{highlight(s=-16t^2+32t)}}}
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b) The ball will be how high above the ground after 1 second?
s(1)=-16(1)^2+32(1)
s(1)=-16(1)+32
s(1)=-16+32
{{{highlight(s(1)=16ft)}}}
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c) How long will it take to hit the ground?
0=-16t^2+32t
0=-16t(t-2)
-16t=0  and t-2=0
-16t/-16=0/-16 and t-2+2=0+2
t=0  and t=2s
t=0 is when it was thrown.
{{{highlight(t=2s)}}} is when it lands.
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d) What is the maximum height of the ball?
The maximum height occurs halfway through the flight of the ball, it lands at t=2 s.  So it's maximum height occurs when t=1.  We found out that it was {{{highlight(s(1)=16ft)}}}  on b.
Happy Calculating!!!