calculus
posted by john on .
A baseball pitcher throws a baseball with an initial velocity of 133 feet per second at an angle of 20° to the hoizontal. The ball leaves the pitcher's hand at a height of 5 feet. Find parametric equations that descrive the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum height? What is the maximum height of the ball?

The equations are:
x = 133 cos(35)* t
y = 5 + 133 sin 35 * t  (1/2) g t^2
It will remain in the air until y = 0. (Solve then y equation for t).
g = 9.8 m/s^2.
Maximum height is attained when
Vy = 133 sin 35  gt = 0.
Solve for t. That is the time when the height is a maximum. Then insert that t into the equation for y(t). That will give you th3e maximum height. 
x = 124..98t; y = 16t^2 + 45.49t + 5
2.729 sec
1.422 sec
4.983 feet
is the answer I'm coming up with ?? 
how do you find the time in the air?? or maximum height??