A baseball player hits a pitch when the ball is 2 feet off the ground. Two hundred feet from the plate, the ball reaches its maximum height of 82 feet. Will the ball clear the home run wall that is 10 feet high and 380 feet from home plate?
The equation below models the height in feet, h, of a softball t seconds after it is hit by a batter.
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Why is the height of the ball a function of time?
Each value of time is associated with exactly one height.
Each height of the ball is associated with exactly one moment in time.
There is exactly one time at which the maximum height is attained.
There is a maximum and minimum height that the ball can reach.
There is a maximum and minimum height that the ball can reach.
Which equation is a function of x?
To determine if the ball will clear the home run wall, we can analyze the trajectory of the ball.
First, let's label the variables given in the problem:
- Initial height (h0) = 2 feet
- Distance from the plate (x) = 200 feet
- Maximum height (h_max) = 82 feet
- Home run wall height (wall_height) = 10 feet
- Distance to the home run wall (wall_distance) = 380 feet
The path of the ball can be described by a parabolic trajectory. The equation for the height (h) of the ball at any horizontal distance (x) is given by:
h = -a(x - p)(x - q) + k
where:
- a is a constant related to the trajectory shape
- p and q are the horizontal distances where the ball reaches its maximum height, forming a symmetric parabola
- k is the maximum height reached by the ball
Our task is to determine if the ball's height clears the home run wall (k + h0 > wall_height) when it reaches the distance (wall_distance).
To find the values of a, p, q, and k, we can use the information given in the problem.
1. Equation for maximum height:
Using the given information, we can set up two equations:
h0 = -a(p - p)(p - q) + k --> equation 1 (at x = 0)
h_max = -a(q - p)(q - q) + k --> equation 2 (at x = 200)
Since q = p (symmetric parabola), we can simplify equation 2:
h_max = -a(q - p)(0) + k
h_max = k
Therefore, we can substitute k with h_max in equation 1:
h0 = -a(q - p)(p - q) + h_max --> equation 3
2. Solving for a:
Since h_max and h0 are given, we have:
82 = -a(200 - p)(p - p) + h_max
82 = -a(200 - p)(0) + h_max
82 = h_max
Now, we have the value for h_max = 82, which allows us to solve for a:
82 = -a(200 - p)(0) + 82
0 = -a(200 - p)
Since the ball reaches its maximum height at x = 200, we have:
0 = -a(200 - 200)
0 = -a(0)
a can be any real number.
After solving for a, we can use the values of p, q, k, and a in the equation:
h = -a(x - p)(x - q) + k
Substituting the known values:
h = -a(x - 200)(x - 200) + 82
Now, we can substitute the distance to the home run wall (wall_distance = 380) into the equation and check if the ball's height (h) clears the wall (k + h0 > wall_height):
h = -a(380 - 200)(380 - 200) + 82
After finding the value of h at x = wall_distance, we can compare it to the wall_height.
Depending on the calculated value of h, we can determine if the ball will clear the home run wall or not.