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The problem provides the function \( y = -16t^2 + 529 \), where \( y \) represents the height of the phone in feet and \( t \) represents the time in seconds after it is dropped.

To find when the phone hits the ground, we set \( y = 0 \) because the phone's height will be 0 feet when it reaches the ground. This gives us the equation:

\[
0 = -16t^2 + 529
\]

### Step 1: Solve for \( t \)

Move \( 529 \) to the other side:

\[
16t^2 = 529
\]

### Step 2: Divide both sides by 16:

\[
t^2 = \frac{529}{16}
\]

\[
t^2 = 33.0625
\]

### Step 3: Take the square root of both sides:

\[
t = \sqrt{33.0625}
\]

\[
t \approx 5.75 \, \text{seconds}
\]

Thus, the phone hits the ground after approximately **5.75 seconds**.

Would you like further details or have any questions about this process?

Here are five related questions:
1. How would the answer change if the initial height were different?
2. Can we solve this problem graphically by plotting the function?
3. What does the coefficient -16 represent in the equation?
4. How can we determine the maximum height of the phone?
5. What happens to the equation if there is an initial upward velocity?

**Tip:** When solving quadratic equations like this one, always check if it can be factored easily before using the quadratic formula or square roots.

While taking a photograph on a balcony, you drop your cell phone. The function y = -16t^2 + 529 represents the height y (in feet) of the phone t seconds after it is dropped. After how many seconds does the phone land on the ground?

Learn how to solve the quadratic equation y = -16t^2 + 529 to determine when an object hits the ground. This problem focuses on key algebra concepts and solving quadratic equations for time, suitable for Grades 8-10.

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