Question
A water balloon is launched up from the ground. After 2.4 seconds, the balloon reaches a maximum height of 28.8 meters. It returns to the ground 4.8 seconds after it is launched.
Let f(x) be the height (in meters) of the balloon r seconds after it is launched. Then, the function fis quadratic. (Its graph is a parabola.) Write an equation for the quadratic function f.
Let f(x) be the height (in meters) of the balloon r seconds after it is launched. Then, the function fis quadratic. (Its graph is a parabola.) Write an equation for the quadratic function f.
Answers
GPT-4o mini
To find the equation of the quadratic function \( f(x) \) that represents the height of the water balloon, we can start with the general form of a quadratic function:
\[
f(t) = at^2 + bt + c
\]
where \( a \), \( b \), and \( c \) are constants.
### Step 1: Determine the vertex
Given that the maximum height of the balloon is 28.8 meters at \( t = 2.4 \) seconds, we know that the vertex of the parabola occurs at \( (2.4, 28.8) \). The vertex form of a quadratic function can be expressed as:
\[
f(t) = a(t - h)^2 + k
\]
where \( (h, k) \) is the vertex of the parabola. So we can substitute:
\[
f(t) = a(t - 2.4)^2 + 28.8
\]
### Step 2: Determine the value of \( a \)
We also know that the balloon returns to the ground (height = 0) at \( t = 4.8 \) seconds. Therefore:
\[
f(4.8) = 0
\]
Substituting \( t = 4.8 \) into our vertex form and setting it equal to 0 gives:
\[
0 = a(4.8 - 2.4)^2 + 28.8
\]
Calculating \( (4.8 - 2.4)^2 \):
\[
(4.8 - 2.4)^2 = (2.4)^2 = 5.76
\]
Now we have:
\[
0 = a(5.76) + 28.8
\]
Solving for \( a \):
\[
a(5.76) = -28.8
\]
\[
a = \frac{-28.8}{5.76} = -5
\]
### Step 3: Write the final equation
Now we can plug \( a \) back into the vertex form:
\[
f(t) = -5(t - 2.4)^2 + 28.8
\]
To convert this into standard form \( f(t) = at^2 + bt + c \), we can expand it:
1. Calculate \( (t - 2.4)^2 \):
\[
(t - 2.4)^2 = t^2 - 4.8t + 5.76
\]
2. Substitute into \( f(t) \):
\[
f(t) = -5(t^2 - 4.8t + 5.76) + 28.8
\]
3. Distribute \(-5\):
\[
f(t) = -5t^2 + 24t - 28.8 + 28.8
\]
4. Simplify:
\[
f(t) = -5t^2 + 24t
\]
So the quadratic function for the height of the balloon as a function of time \( t \) is:
\[
\boxed{f(t) = -5t^2 + 24t}
\]
\[
f(t) = at^2 + bt + c
\]
where \( a \), \( b \), and \( c \) are constants.
### Step 1: Determine the vertex
Given that the maximum height of the balloon is 28.8 meters at \( t = 2.4 \) seconds, we know that the vertex of the parabola occurs at \( (2.4, 28.8) \). The vertex form of a quadratic function can be expressed as:
\[
f(t) = a(t - h)^2 + k
\]
where \( (h, k) \) is the vertex of the parabola. So we can substitute:
\[
f(t) = a(t - 2.4)^2 + 28.8
\]
### Step 2: Determine the value of \( a \)
We also know that the balloon returns to the ground (height = 0) at \( t = 4.8 \) seconds. Therefore:
\[
f(4.8) = 0
\]
Substituting \( t = 4.8 \) into our vertex form and setting it equal to 0 gives:
\[
0 = a(4.8 - 2.4)^2 + 28.8
\]
Calculating \( (4.8 - 2.4)^2 \):
\[
(4.8 - 2.4)^2 = (2.4)^2 = 5.76
\]
Now we have:
\[
0 = a(5.76) + 28.8
\]
Solving for \( a \):
\[
a(5.76) = -28.8
\]
\[
a = \frac{-28.8}{5.76} = -5
\]
### Step 3: Write the final equation
Now we can plug \( a \) back into the vertex form:
\[
f(t) = -5(t - 2.4)^2 + 28.8
\]
To convert this into standard form \( f(t) = at^2 + bt + c \), we can expand it:
1. Calculate \( (t - 2.4)^2 \):
\[
(t - 2.4)^2 = t^2 - 4.8t + 5.76
\]
2. Substitute into \( f(t) \):
\[
f(t) = -5(t^2 - 4.8t + 5.76) + 28.8
\]
3. Distribute \(-5\):
\[
f(t) = -5t^2 + 24t - 28.8 + 28.8
\]
4. Simplify:
\[
f(t) = -5t^2 + 24t
\]
So the quadratic function for the height of the balloon as a function of time \( t \) is:
\[
\boxed{f(t) = -5t^2 + 24t}
\]