To find the quadratic function representing the height of the soccer ball in relation to time, we can use the general form of a quadratic function:
\[ h(t) = at^2 + bt + c \]
where \( h(t) \) is the height, \( t \) is time, \( a \), \( b \), and \( c \) are coefficients we need to determine.
From the data provided:
- At \( t = 0 \) seconds, \( h(0) = 0 \) ft.
- At \( t = 0.2 \) seconds, \( h(0.2) = 4 \) ft.
- At \( t = 0.6 \) seconds, \( h(0.6) = 12 \) ft.
- At \( t = 1 \) second, \( h(1) = 18 \) ft.
- At \( t = 1.5 \) seconds, \( h(1.5) = 20 \) ft.
- At \( t = 2 \) seconds, \( h(2) = 16 \) ft.
With the height \( h(0) = 0 \) at \( t = 0 \), this gives us the value of \( c = 0 \), so we can simplify our function to:
\[ h(t) = at^2 + bt \]
We can use the other points to set up a system of equations:
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For \( t = 0.2 \): \[ h(0.2) = a(0.2)^2 + b(0.2) = 4 \] \[ 0.04a + 0.2b = 4 \quad \text{(1)} \]
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For \( t = 0.6 \): \[ h(0.6) = a(0.6)^2 + b(0.6) = 12 \] \[ 0.36a + 0.6b = 12 \quad \text{(2)} \]
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For \( t = 1 \): \[ h(1) = a(1)^2 + b(1) = 18 \] \[ a + b = 18 \quad \text{(3)} \]
Now we have a system of three equations. We can solve equations (1), (2), and (3).
From equation (3), we can express \( b \) in terms of \( a \):
\[ b = 18 - a \]
Substituting \( b \) into equations (1) and (2):
Substituting into equation (1):
\[ 0.04a + 0.2(18 - a) = 4 \] \[ 0.04a + 3.6 - 0.2a = 4 \] \[ -0.16a + 3.6 = 4 \] \[ -0.16a = 0.4 \] \[ a = -\frac{0.4}{0.16} = -2.5 \]
Now substituting \( a = -2.5 \) back into the equation for \( b \):
\[ b = 18 - (-2.5) = 18 + 2.5 = 20.5 \]
Thus, we have:
- \( a = -2.5 \)
- \( b = 20.5 \)
- \( c = 0 \)
Finally, rounding \( b = 20.5 \) to the nearest hundredth,
the value of \( b \) is:
\[ \boxed{20.50} \]