the height of the ball 10 feet before hitting the ground is approximately 15.48 feet.
a) The equation to model the height of the ball (H) based on the horizontal distance (x) can be represented by a quadratic function in standard form:
H(x) = -ax^2 + bx + c
To find the values of the coefficients a, b, and c, we can use the given information:
At the starting point, x = 0 and H(0) = 6, so c = 6.
At the maximum height, x = 30 and H(30) = 39, so:
39 = -a(30)^2 + b(30) + 6
39 = -900a + 30b + 6
33 = 30b - 900a
b = (33 + 900a)/30
To find the value of a, we can use the fact that the ball reaches its maximum height halfway through its horizontal distance:
x = 30, H(x) = 39, and H'(x) = 0
H(x) = -ax^2 + bx + c
39 = -900a + 30b + 6
0 = -60a + b
b = 60a
Substituting b = 60a into the equation from the previous step, we get:
b = (33 + 900a)/30 = 60a
33 = 1740a
a = 33/1740 = 0.01897 (rounded to 5 decimal places)
Therefore, the equation to model the height of the ball is:
H(x) = -0.01897x^2 + 1.1381x + 6
b) To find the height of the ball 10 feet before hitting the ground, we can plug in x = 50 (since the ball travels 60 feet in total) and solve for H(x):
H(x) = -0.01897x^2 + 1.1381x + 6
H(50) = -0.01897(50)^2 + 1.1381(50) + 6
H(50) = -47.425 + 56.905 + 6
H(50) = 15.48
Therefore, the height of the ball 10 feet before hitting the ground is approximately 15.48 feet.