Question 52429
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?
Answer: s(t) = -162 + 32t

b) 	The ball will be how high above the ground after 1 second?
Answer:	16 feet
Show work in this space.	

s(t) = -162 + 32t
s(1) = -16*(1)2 + 32*(1)
s(1) = -16 + 32
s(1) = 16 feet

c) 	How long will it take to hit the ground?
Answer:	2 seconds
Show work in this space. 

s(t) = -162 + 32t
t(-16t + 3) = 0
t=0 & -16t+ 32 = 0

-16t+ 32 = 0
-16t + 32 - 32 = 0 -32
-16t = -32
-16t/-16 = -32/-16
t = 2

d) 	What is the maximum height of the ball?  What time will the maximum height be attained?
Answer:Maximum height is 16 feet. The maximum height will be attained in 1 second.

Show work in this space.	
t = -b/2*9
	t = -32/2*(-16)
	t = 1 second