If you are standing near the edge of the top of a 200 feet building and throw a ball vertically upward it will be modeled by the function s(t)=-16t^2+64t+200 where s(t) is the ball's height above ground in feet and t is seconds after the ball was thrown.

Question

If you are standing near the edge of the top of a 200 feet building and throw a ball vertically upward it will be modeled by the function s(t)=-16t^2+64t+200 where s(t) is the ball's height above ground in feet and t is seconds after the ball was thrown. 
A) When does the ball reach its maximum height and what is the maximum height? 
T= -b/2a
T= -64/-2(-16)
T= 2 seconds 
How do I find the maximum height?
B) When does the ball hit the ground? (Round to the nearest tenth of a second)
Can you please help me with this question? 
C) Find s(0) and describe what it represents?
Can you please help me with this question?
D) How would I graph the quadratic function?

Reiny · Answer

to find max height, plug your t = 2 into the original equation
s(2) = -16(4) + 64(2) + 200
= .....

at ground, s(t) = 0
-16t^2 + 64t + 200 = 0
4t^2 - 16t - 50 = 0
2t^2 - 8t - 25 = 0
use the quadratic formula:
t = (8 ± √264)/4
= appr 6.062 or a negative, which we will reject

verification:
http://www.wolframalpha.com/input/?i=s%28t%29%3D-16t%5E2%2B64t%2B200+