A cannon ball is launched into the air with an upward velocity of 227 feet per second, from a 15-foot tall cannon. The height ℎ of the cannon ball after 𝑡 seconds can be found using the equation ℎ=-16𝑡2+227𝑡+15.

Approximately how long will it take for the cannon ball to be 426 feet high? Round answers to the nearest tenth if necessary.
 units: 



How long long will it take to hit the ground?
 units: 

To solve these problems, we'll use the given equation for the height of the cannonball:

\[ h = -16t^2 + 227t + 15 \]

### 1. Time to reach 426 feet

We need to find the time \( t \) when the height \( h \) is 426 feet. Set \( h = 426 \) and solve for \( t \):

\[ 426 = -16t^2 + 227t + 15 \]

Rearrange this into a standard quadratic equation form:

\[ -16t^2 + 227t + 15 - 426 = 0 \]
\[ -16t^2 + 227t - 411 = 0 \]

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 227 \), and \( c = -411 \):

First, calculate the discriminant:

\[ b^2 - 4ac = 227^2 - 4(-16)(-411) \]
\[ b^2 - 4ac = 51529 - 4 \cdot 16 \cdot 411 \]
\[ b^2 - 4ac = 51529 - 26256 \]
\[ b^2 - 4ac = 25273 \]

Now, solve for \( t \):

\[ t = \frac{-227 \pm \sqrt{25273}}{2(-16)} \]
\[ t = \frac{-227 \pm 159.01}{-32} \]

This gives us two possible solutions:

\[ t_1 = \frac{-227 + 159.01}{-32} \approx \frac{-67.99}{-32} \approx 2.1 \]
\[ t_2 = \frac{-227 - 159.01}{-32} \approx \frac{-386.01}{-32} \approx 12.1 \]

So, the cannonball will reach a height of 426 feet at approximately 2.1 seconds and again at 12.1 seconds.

### 2. Time to hit the ground

To find out how long it takes for the cannonball to hit the ground, we set \( h = 0 \):

\[ 0 = -16t^2 + 227t + 15 \]

Again, we use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Calculate the discriminant:

\[ b^2 - 4ac = 227^2 - 4(-16)(15) \]
\[ b^2 - 4ac = 51529 + 960 \]
\[ b^2 - 4ac = 52489 \]

Now, solve for \( t \):

\[ t = \frac{-227 \pm \sqrt{52489}}{2(-16)} \]
\[ t = \frac{-227 \pm 229.01}{-32} \]

This gives us two possible solutions:

\[ t_1 = \frac{-227 + 229.01}{-32} \approx \frac{2.01}{-32} \approx -0.1 \] (Not physically meaningful as time cannot be negative)
\[ t_2 = \frac{-227 - 229.01}{-32} \approx \frac{-456.01}{-32} \approx 14.3 \]

So, the cannonball will hit the ground after approximately 14.3 seconds.

### Summary

1. Time to reach 426 feet:
   - \( \approx 2.1 \) seconds
   - \( \approx 12.1 \) seconds
2. Time to hit the ground:
   - \( \approx 14.3 \) seconds

Would you like details or have any questions?

Here are some more questions you might want to ask:
1. How do you use the quadratic formula to solve equations?
2. What does the discriminant tell us about the solutions to a quadratic equation?
3. How do you interpret the physical meaning of negative time solutions?
4. Can you explain the process of solving quadratic equations step by step?
5. How do you determine when a projectile reaches its maximum height?

**Tip:** Always check the physical context of the problem when interpreting solutions to quadratic equations, especially when dealing with time.

Explore how to calculate the time for a cannonball to reach 426 feet and hit the ground using quadratic equations. Detailed solutions and explanations included. Suitable for high school students.

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