A shot put champion releases the shot from 6ft above the ground. The path of the shot is modeled by:

h(d)=-0.02d^2+1.11d+c.
How far does the shot travel before it hits the ground?

h is approximate height (in feet), d is the horizontal distance traveled (in feet), and c is the initial height (in feet).

To find out how far the shot travels before it hits the ground, we need to determine the value of d when h(d) equals zero. When h(d) equals zero, it means that the height of the shot is at ground level.

Given that the path of the shot is modeled by the equation h(d) = -0.02d^2 + 1.11d + c, we can substitute h(d) with zero:

0 = -0.02d^2 + 1.11d + c

Since c represents the initial height, which is given as 6ft, we can substitute c with 6:

0 = -0.02d^2 + 1.11d + 6

Now we have a quadratic equation in the form of Ax^2 + Bx + C = 0, where A = -0.02, B = 1.11, and C = 6.

To solve this quadratic equation, you can use the quadratic formula:

x = (-B ± √(B^2 - 4AC)) / (2A)

Substituting the values of A, B, and C into the formula, we have:

d = (-1.11 ± √(1.11^2 - 4(-0.02)(6))) / (2(-0.02))

Simplifying further:

d = (-1.11 ± √(1.2321 + 0.48)) / (-0.04)

d = (-1.11 ± √1.7121) / (-0.04)

Now, we can calculate the two possible values of d by using both the positive (+) and negative (-) square roots:

d₁ = (-1.11 + √1.7121) / (-0.04)
d₂ = (-1.11 - √1.7121) / (-0.04)

After calculating these expressions, we can find the real values of d that represents the distance the shot travels before hitting the ground.