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. How many seconds after the projectile is launched will it hit the ground?

Answer

To find out how many seconds after the projectile is launched it will hit the ground, we need to determine when the height \( h(t) \) is equal to zero. The given equation for height is:

\[ h(t) = -4.9t^2 + 88.2t \]

### Step 1: Set the height equation to zero
To find when the projectile hits the ground, we set \( h(t) = 0 \):

\[ -4.9t^2 + 88.2t = 0 \]

### Step 2: Factor out common terms
We can factor out \( t \) from the equation:

\[ t(-4.9t + 88.2) = 0 \]

### Step 3: Solve for \( t \)
This gives us two possible solutions:

1. \( t = 0 \)
2. \( -4.9t + 88.2 = 0 \)

For the second equation, we can solve for \( t \):

\[ -4.9t + 88.2 = 0 \implies 4.9t = 88.2 \implies t = \frac{88.2}{4.9} \]

### Step 4: Calculate \( t \)
Now we perform the division:

\[ t = \frac{88.2}{4.9} \approx 18 \]

### Conclusion
Thus, the projectile will hit the ground approximately **18 seconds** after it is launched.

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