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?

calculus 
drwls,
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. 
calculus 
john,
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 ?? 
calculus 
Carlos Garcia,
how do you find the time in the air?? or maximum height??