Question

A ball is thrown up into the air from a height of 17.6 meters above the ground. After 1.8 seconds, the ball reaches a maximum height of 33.8 meters. It then begins to fall and hits the ground 4.4 seconds after it is thrown.
Let f(x) be the height (in meters) of the ball r seconds after it is thrown. Then, the function fis quadratic.
(Its graph is a parabola.) Write an equation for the quadratic function f.

Answer 1

To find the quadratic function \( f(r) \) that describes the height of the ball at time \( r \), we can use the standard form of a quadratic equation:

\[
f(r) = ar^2 + br + c
\]

where \( a \), \( b \), and \( c \) are constants to be determined.

### Step 1: Identify the given points

From the problem, we know:

1. The initial height at \( r = 0 \) seconds (when thrown) is \( f(0) = 17.6 \) meters. Therefore, \( c = 17.6 \).
2. At \( r = 1.8 \) seconds, the ball reaches a maximum height of \( 33.8 \) meters, so \( f(1.8) = 33.8 \).
3. The ball hits the ground at \( r = 4.4 \) seconds, hence \( f(4.4) = 0 \).

### Step 2: Build equations based on the points

We can set up the following equations based on the points we have:

1. \( f(0) = c = 17.6 \)
2. At \( r = 1.8 \):
\[
f(1.8) = a(1.8)^2 + b(1.8) + 17.6 = 33.8
\]
Simplifying this, we have:
\[
3.24a + 1.8b + 17.6 = 33.8
\]
\[
3.24a + 1.8b = 16.2 \quad \text{(Equation 1)}
\]

3. At \( r = 4.4 \):
\[
f(4.4) = a(4.4)^2 + b(4.4) + 17.6 = 0
\]
Simplifying this, we have:
\[
19.36a + 4.4b + 17.6 = 0
\]
\[
19.36a + 4.4b = -17.6 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations

We now solve the two equations simultaneously:

**Equation 1:**
\[
3.24a + 1.8b = 16.2 \quad \text{(1)}
\]

**Equation 2:**
\[
19.36a + 4.4b = -17.6 \quad \text{(2)}
\]

To eliminate one of the variables, we can multiply Equation 1 by \( \frac{4.4}{1.8} \) to align the coefficients of \( b \):

\[
\frac{4.4}{1.8} \cdot (3.24a + 1.8b) = \frac{4.4}{1.8} \cdot 16.2
\]

Calculating this gives:
\[
8a + 4.4b = 36.4 \quad \text{(3)}
\]

Now we can subtract Equation 2 from Equation 3:

\[
(8a + 4.4b) - (19.36a + 4.4b) = 36.4 + 17.6
\]
\[
-11.36a = 54
\]
\[
a = -\frac{54}{11.36} \approx -4.75
\]

Now substitute the value of \( a \) back into Equation 1 to find \( b \):

\[
3.24(-4.75) + 1.8b = 16.2
\]
\[
-15.39 + 1.8b = 16.2
\]
\[
1.8b = 31.59
\]
\[
b = \frac{31.59}{1.8} \approx 17.54
\]

### Step 4: Write the final equation

Now we have \( a \approx -4.75 \), \( b \approx 17.54 \), and \( c = 17.6 \). Therefore, the quadratic function can be approximated as:

\[
f(r) = -4.75r^2 + 17.54r + 17.6
\]

Thus the final equation for the height of the ball at time \( r \) seconds is:

\[
f(r) \approx -4.75r^2 + 17.54r + 17.6
\]