Posted by **john** on Sunday, December 2, 2007 at 6:53pm.

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**, Sunday, December 2, 2007 at 7:40pm
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 -
**Carlos Garcia**, Wednesday, April 20, 2016 at 11:20am
how do you find the time in the air?? or maximum height??

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