Question 55874
Hi Tine,
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?
Answer:{{{s=-16t^2+32t}}}  
v0=32 and s0=0
{{{s=-16t^2+(32)t+0}}}
{{{s=-16t^2+32t}}}


b) 	The ball will be how high above the ground after 1 second?
Answer:  16 ft
Show work in this space.  
{{{s(1)=-16(1)^2+32(1)}}}
s(1)=-16+32
s(1)=16 ft


c) 	How long will it take to hit the ground?
Answer:  t=2s
Show work in this space.  
{{{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=-2
t=0 s and t=2 s
At t=0s the ball was thrown and at t=2s the ball hit the ground.



d) 	What is the maximum height of the ball?
Answer:  16 ft.
Show work in this space.  
The maximum height is the vertex of the parabola.  We find the t value of the vertex by the formula:{{{highlight(t=-b/2a)}}} 
{{{t=-(32)/(2(-16))}}}
{{{t=-32/-32}}}
t=1  at t=1s the ball reaches its maximum height.
{{{s(1)=-16(1)^2+32(1)}}}
{{{s(1)=-16+32}}}
s(1)=16 ft