The height in feet of a projectile with an initial velocity of 64 feet per second and an initial height of 192 feet is a function of time in seconds given by
h(t) = −16t2 + 64t + 192.
(a) Find the maximum height of the projectile.
 ft

(b) Find the time t when the projectile achieves its maximum height.
t = 
 sec

(c) Find the time t when the projectile has a height of 0 feet.
t = 
 sec

We are given the height function of a projectile:

\[
h(t) = -16t^2 + 64t + 192
\]

### (a) Finding the maximum height:

This is a quadratic function, and the maximum height occurs at the vertex. The time \(t\) at which the vertex occurs can be found using the formula:

\[
t = \frac{-b}{2a}
\]

Where \(a = -16\) and \(b = 64\). Plugging in these values:

\[
t = \frac{-64}{2(-16)} = \frac{-64}{-32} = 2 \, \text{seconds}
\]

Now, substitute \(t = 2\) into the height function to find the maximum height:

\[
h(2) = -16(2)^2 + 64(2) + 192 = -16(4) + 128 + 192 = -64 + 128 + 192 = 256 \, \text{feet}
\]

So, the maximum height is **256 feet**.

### (b) The time when the projectile achieves its maximum height:

From the previous calculation, the maximum height occurs at **\(t = 2\) seconds**.

### (c) Finding the time when the projectile has a height of 0 feet:

To find when the projectile reaches a height of 0 feet, we set \(h(t) = 0\):

\[
-16t^2 + 64t + 192 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = -16\), \(b = 64\), and \(c = 192\). Plugging these values into the quadratic formula:

\[
t = \frac{-64 \pm \sqrt{64^2 - 4(-16)(192)}}{2(-16)}
\]
\[
t = \frac{-64 \pm \sqrt{4096 + 12288}}{-32}
\]
\[
t = \frac{-64 \pm \sqrt{16384}}{-32}
\]
\[
t = \frac{-64 \pm 128}{-32}
\]

Now, solving for both possible values of \(t\):

1. \(t = \frac{-64 + 128}{-32} = \frac{64}{-32} = -2\) (This is not physically meaningful since time cannot be negative).
2. \(t = \frac{-64 - 128}{-32} = \frac{-192}{-32} = 6\).

Thus, the time when the projectile hits the ground is **6 seconds**.

### Summary:

- (a) Maximum height = **256 feet**
- (b) Time to reach maximum height = **2 seconds**
- (c) Time when the projectile hits the ground = **6 seconds**

Would you like more details on any part of this problem?

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Here are 5 related questions for further practice:
1. What would happen to the maximum height if the initial velocity were increased to 80 feet per second?
2. How would the equation change if the initial height were 100 feet?
3. Can you determine when the projectile reaches a height of 150 feet?
4. What is the total flight time of the projectile if the initial velocity was halved?
5. What is the velocity of the projectile at \(t = 3\) seconds?

**Tip:** The quadratic formula can always help solve for time when height or position is zero!

The height in feet of a projectile with an initial velocity of 64 feet per second and an initial height of 192 feet is a function of time in seconds given by h(t) = −16t^2 + 64t + 192.
(a) Find the maximum height of the projectile.
(b) Find the time t when the projectile achieves its maximum height.
(c) Find the time t when the projectile has a height of 0 feet.

Learn how to solve a projectile motion problem using quadratic equations. This example walks through finding the maximum height and the time when the projectile hits the ground, with detailed step-by-step solutions. Suitable for students in Grades 9-10.

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