Posted by Abbey on .
Projectile motion: Let's suppose you throw a ball straight up with an initial speed of 50 feet per second from a height of feet.
a) Find the equation that describes the motion as a function of time.
My answer was x = 0 and y=16t^2+50t+6
b) How long is the ball in the air?
This is the part I need help on. Can I put the y equation into my graphing calculator to solve for t?
c) Determine when the ball is at maximum height. Find the maximum height.

Math(Please help) 
Damon,
when is y = 0 (hits ground)?
0 = 16 t^2 50 t  6 (I multiplied both sides by 1)
0 = 8 t^2 25 t 3
solve quadratic for t
t = [ 25 +/ sqrt (625 + 96) ] / 32
t = [25 +/ 26.8514]/16
use the + root the  root is before you threw it
t = 3.24 seconds
max height when velocity = 0
V = Vo 32 t
32 t = 50
t = 1.56 seconds
height = 6 + 50 (1.56)  16 (1.56^2) 
Math(Please help) 
Abbey,
Thank You!!