Questions 37 and 38 refer to the following information.
`h(t)=-4.9t^(2)+88.2t`
When a projectile is launched from ground level, the equation above gives the number of meters in its height, h, after t seconds have elapsed.
Q. What is the maximum height the projectile reaches, correct to the nearest meter?

Answer

To find the maximum height of the projectile described by the equation \( h(t) = -4.9t^2 + 88.2t \), we will follow these steps:

### Step 1: Identify the function
The height of the projectile is given by the function:
\[ h(t) = -4.9t^2 + 88.2t \]

### Step 2: Differentiate the function
To find the maximum height, we need to find the critical points by differentiating \( h(t) \) with respect to \( t \):
\[ \frac{dh}{dt} = \frac{d}{dt}(-4.9t^2 + 88.2t) \]
Using the power rule for differentiation, we get:
\[ \frac{dh}{dt} = -9.8t + 88.2 \]

### Step 3: Set the derivative equal to zero
To find the time \( t \) at which the maximum height occurs, we set the derivative equal to zero:
\[ -9.8t + 88.2 = 0 \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:
\[ 9.8t = 88.2 \]
Now, divide both sides by 9.8:
\[ t = \frac{88.2}{9.8} = 9 \]

### Step 5: Substitute \( t \) back into the original function
Now that we have \( t = 9 \), we substitute this value back into the original height function to find the maximum height:
\[ h(9) = -4.9(9^2) + 88.2(9) \]
Calculating \( 9^2 \):
\[ 9^2 = 81 \]
Now substitute:
\[ h(9) = -4.9(81) + 88.2(9) \]
Calculating each term:
\[ -4.9 \times 81 = -396.9 \]
\[ 88.2 \times 9 = 793.8 \]
Now combine these results:
\[ h(9) = -396.9 + 793.8 = 396.9 \]

### Step 6: Round to the nearest meter
The maximum height, rounded to the nearest meter, is:
\[ \text{Maximum Height} \approx 397 \text{ meters} \]

### Final Answer
The maximum height the projectile reaches is **397 meters**.

---