To rewrite the height function \( h(t) = -5t^2 + 5t + 10 \) in a different form, we will complete the square.
- Completing the Square:
First, factor out \(-5\) from the first two terms:
\[ h(t) = -5(t^2 - t) + 10 \]
Next, take half of the coefficient of \(t\) (which is \(-1\)), square it, and add and subtract this value inside the parentheses. Half of \(-1\) is \(-\frac{1}{2}\), and squaring it gives \(\frac{1}{4}\).
\[ h(t) = -5\left(t^2 - t + \frac{1}{4} - \frac{1}{4}\right) + 10 \]
Now rewrite it as:
\[ h(t) = -5\left((t - \frac{1}{2})^2 - \frac{1}{4}\right) + 10 \]
Distributing \(-5\):
\[ h(t) = -5(t - \frac{1}{2})^2 + \frac{5}{4} + 10 \]
Convert \(10\) to quarters:
\[ 10 = \frac{40}{4} \]
Adding the constants gives:
\[ h(t) = -5(t - \frac{1}{2})^2 + \frac{45}{4} \]
Thus, the vertex form of the function is:
\[ h(t) = -5\left(t - \frac{1}{2}\right)^2 + \frac{45}{4} \]
- Finding when the brush hits the ground:
To find when the brush hits the ground, we set \(h(t) = 0\):
\[ -5(t - \frac{1}{2})^2 + \frac{45}{4} = 0 \]
Rearranging gives:
\[ -5(t - \frac{1}{2})^2 = -\frac{45}{4} \]
Dividing both sides by \(-5\):
\[ (t - \frac{1}{2})^2 = \frac{45}{20} = \frac{9}{4} \]
Now, take the square root of both sides:
\[ t - \frac{1}{2} = \pm\frac{3}{2} \]
Solving for \(t\):
- \(t - \frac{1}{2} = \frac{3}{2}\):
\[ t = 2 \]
- \(t - \frac{1}{2} = -\frac{3}{2}\):
\[ t = -1 \quad (\text{not a valid solution for time}) \]
Thus, the brush hits the ground 2 seconds after being thrown.
Final Answer:
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\(h(t) = -5\left(t - \frac{1}{2}\right)^2 + \frac{45}{4}\)
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The brush hits the ground after \(2\) seconds.