Question 66387
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 + vot + so 
•	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?
s(t)=-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)=16 ft high
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c) 	How long will it take to hit the ground?
When it hits the ground its height will be zero, so:
0=-16t^2+32t
-16t(t-2)=0
t=0 or t=2
t=0 means the ball starts on the ground
t=32 means the ball is on the ground again after 2 seconds
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d) 	What is the maximum height of the ball?
s(t)=-16t^2+32t 
This is a quadratic with a=-16 and b=32
The vertex of the parabola occurs at t=-b/2a=-(32)/(-32)=1
So the top of the arc occurs at time t=1 sec.
The height at that time is 16 feet as was shown in part "b".
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Cheers,
Stan H.