a baseball player hits sa 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?

Are we to assume that the path of the ball follows a quadratic function ?

If so, then (200,82) would be the vertex.

let the equation be y = a(x-200)^2 + 82

when x = 0, y = 2
2 = a(-200)^2 + 82
a = -1/500

so y = (-1/500)(x-200)^2 + 82

when x = 380
y = (-1/500)(380-200)^2 + 82 = 17.2
which is more than 10 feet

so , yes, it will clear the home run wall

To determine whether the ball will clear the home run wall, we need to find out the height of the ball when it reaches the home run wall location.

First, let's define the variables:
- Initial height of the ball (when hit): h_initial = 2 feet
- Maximum height reached by the ball: h_max = 82 feet
- Distance from home plate to the ball at maximum height: d_max = 200 feet
- Distance from home plate to the home run wall: d_wall = 380 feet
- Height of the home run wall: h_wall = 10 feet

We can use a projectile motion equation to find the height of the ball at any given distance. The equation for the height of a projectile is given by:
h = h_initial + (d - d_initial) * tan(angle) - (g * (d - d_initial)^2) / (2 * (v_initial * cos(angle))^2)

In this equation:
- h is the height of the ball at distance d
- d is the distance from home plate
- d_initial is the initial distance from home plate (when ball hit)
- angle is the launch angle of the ball
- g is the acceleration due to gravity (9.8 m/s^2 or 32.2 ft/s^2)
- v_initial is the initial velocity of the ball

Since we don't know the launch angle or initial velocity of the ball, we need to solve for them using the given information. We can assume the angle is 45 degrees for simplicity.

To find the initial velocity, we can use the fact that the range (horizontal distance) of a projectile is given by:
range = v_initial^2 * sin(2*angle) / g

The range of the ball when it reaches its maximum height is the distance from home plate to the maximum height location (200ft):
200 = v_initial^2 * sin(90 degrees) / g [since sin(90 degrees) = 1]
200 = v_initial^2 / g

Rearranging this equation, we can solve for v_initial:
v_initial^2 = 200 * g
v_initial = sqrt(200 * g)

Now that we have the initial velocity, we can calculate the height of the ball when it reaches the home run wall location (380ft) using the projectile motion equation:

h_wall = h_initial + (d_wall - d_initial) * tan(angle) - (g * (d_wall - d_initial)^2) / (2 * (v_initial * cos(angle))^2)

Using the values:
h_initial = 2 feet
d_wall = 380 feet
g = 32.2 ft/s^2 (approximate value),
angle = 45 degrees,
v_initial = sqrt(200 * 32.2) feet/s (approximate value)

We can substitute these values and calculate the height of the ball at the home run wall location.