Question 75952
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 ½g, the gravitational pull due to gravity (measured in feet per second 2). 
•	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?
v0=32 ft/s; s0=0 ft
s=-16t^2+32t+0
s=-16t^2+32t
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b) The ball will be how high above the ground after 1 second?
s=-16(1)^2+32(1)
s=-16(1)+32
s=-16+32
s=16 ft.
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c) How long will it take to hit the ground?
0=-16t^2+32t
0=-16t(t-2)
-16t=0 or t-2=0
-16t/-16=0/-16 or t-2+2=0+2
t=0 or t=2
When t=0 the ball is shot up, the ball lands at t=2 s.
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d) What is the maximum height of the ball?
The ball reaches its maximum height at the vertex of the parabola represented by the quadratic equation.  There are a couple of ways of finding that.  This is just one way.
You can use the formual {{{highlight(t=-b/2a)}}}, to find the value of t and then substitute that value into the equation to find the height s.
a=-16 and b=32
{{{t=-32/(2(-16))}}}
{{{t=-32/-32}}}
t=1 s
We found the height at t=1 s for b.
The maximum height is s=16 ft.
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Happy Calculating!!!!