Let's solve this problem step by step:
(a) Find parametric equations that describe the motion of the ball as a function of time.
The motion of the ball can be described using parametric equations that relate the horizontal and vertical positions of the ball (x and y) as functions of time (t). We'll use the following equations:
1. Horizontal motion:
x(t) = initial horizontal position (constant)
2. Vertical motion:
y(t) = initial vertical position + initial vertical velocity * t - (1/2) * acceleration due to gravity * t^2
Given the information:
- Initial speed (vertical velocity): 59 feet per second (upward, so it's positive)
- Initial height: 6 feet
- Acceleration due to gravity: approximately 32.2 feet per second squared (downward, so it's negative)
We can write the parametric equations as follows:
Horizontal motion (x(t)):
x(t) = 0 (assuming the initial position is at the origin)
Vertical motion (y(t)):
y(t) = 6 + 59t - (1/2) * 32.2 * t^2
(b) How long is the ball in the air?
The ball is in the air from the moment it is thrown until it hits the ground. To find the time it's in the air, we need to determine when y(t) becomes zero (the moment it hits the ground). So, we set y(t) = 0 and solve for t:
0 = 6 + 59t - (1/2) * 32.2 * t^2
This is a quadratic equation in t. Let's solve it:
(1/2) * 32.2 * t^2 - 59t - 6 = 0
Now, you can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation:
- a = (1/2) * 32.2
- b = -59
- c = -6
Calculate the two possible values of t using the quadratic formula. One of them will correspond to when the ball is thrown, and the other will correspond to when it hits the ground. The positive value of t will represent the time the ball is in the air.
Once you find the positive value of t, you'll have the time the ball is in the air.