calculus

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?

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asked by john
  1. 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.

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    posted by drwls
  2. 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 ??

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    posted by john
  3. how do you find the time in the air?? or maximum height??

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