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??

To find the parametric equations describing the motion of the ball, we need to break down the initial velocity into horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to gravity.

Let's break down the initial velocity:
- The horizontal component can be found using the formula: Vx = initial velocity * cos(theta)
Vx = 133 ft/s * cos(20°)

- The vertical component can be found using the formula: Vy = initial velocity * sin(theta)
Vy = 133 ft/s * sin(20°)

The horizontal distance covered by the ball can be calculated using the equation: x = Vx * t, where t represents time. Since Vx remains constant, the equation becomes: x = Vx * t.

The vertical distance covered by the ball can be calculated using the equation: y = Vy * t + (1/2) * g * t^2, where g represents the acceleration due to gravity (32.2 ft/s^2). Since the ball starts at a height of 5 feet, the equation becomes: y = Vy * t + (1/2) * g * t^2 + 5.

Now we have the parametric equations that describe the motion of the ball:
x = Vx * t
y = Vy * t + (1/2) * g * t^2 + 5

To find how long the ball is in the air, we can set the vertical distance (y) to zero and solve for t:
0 = Vy * t + (1/2) * g * t^2 + 5
This gives us a quadratic equation that we can solve to find the time(s) when the ball is at ground level.

To find when the ball is at its maximum height, we can find the time when the vertical velocity (Vy) is zero. We can use the equation:
0 = Vy + g * t
Solve for t to find the time when the ball reaches its maximum height.

To find the maximum height of the ball, substitute the time obtained above into the equation for y and solve for y. This will give the maximum height the ball reaches.

Now you have the parametric equations for the motion of the ball, the time the ball is in the air, the time the ball reaches its maximum height, and the maximum height of the ball.