A projectile is fired straight up with a velocity of

320ft/s. Its altitude​ (height), h, after t seconds is given by ​h(t)=−16t^2+320t

**a.** What is the maximum height of the​ projectile?

**b.** When does the projectile hit the​ ground?

Let's solve the problem using the given formula for the projectile's height:
\[
h(t) = -16t^2 + 320t
\]
where \( h(t) \) represents the height of the projectile in feet after \( t \) seconds.

### Part (a): Maximum Height of the Projectile

To find the maximum height, we need to determine the vertex of the quadratic equation. Since this quadratic is in the form \( ax^2 + bx + c \), the time at which the maximum height is reached can be found using the formula:
\[
t_{\text{max}} = \frac{-b}{2a}
\]
Here, \( a = -16 \) and \( b = 320 \). Plugging these values in:
\[
t_{\text{max}} = \frac{-320}{2 \cdot (-16)} = \frac{-320}{-32} = 10 \text{ seconds}
\]

Now, substitute \( t = 10 \) seconds into the equation for \( h(t) \) to find the maximum height:
\[
h(10) = -16(10)^2 + 320(10)
\]
\[
h(10) = -16(100) + 3200
\]
\[
h(10) = -1600 + 3200 = 1600 \text{ feet}
\]

So, the maximum height of the projectile is **1600 feet**.

### Part (b): Time When the Projectile Hits the Ground

The projectile hits the ground when its height is zero, i.e., when \( h(t) = 0 \). We need to solve the equation:
\[
-16t^2 + 320t = 0
\]

Factoring out the common term \( t \):
\[
t(-16t + 320) = 0
\]

This gives us two solutions:
\[
t = 0 \quad \text{or} \quad -16t + 320 = 0
\]
Solving the second equation:
\[
-16t + 320 = 0
\]
\[
t = \frac{320}{16} = 20 \text{ seconds}
\]

So, the projectile hits the ground after **20 seconds**.

### Final Answers:
- **(a)** The maximum height of the projectile is **1600 feet**.
- **(b)** The projectile hits the ground after **20 seconds**.

Would you like more details or have any questions?

Here are some related questions to expand your understanding:

1. How would the maximum height change if the initial velocity were increased?
2. What would happen to the time when the projectile hits the ground if gravity's effect increased?
3. How does the value of the coefficient of the \(t^2\) term relate to the concavity of the parabola?
4. Can the maximum height be determined directly without using the vertex formula?
5. How do you find the initial velocity if you know the time to reach the maximum height?

**Tip:** When dealing with projectile motion, the coefficient of the \(t^2\) term represents the acceleration due to gravity, which is always negative when the object is moving upwards.

A projectile is fired straight up with a velocity of 320ft/s. Its altitude (height), h, after t seconds is given by h(t) = -16t^2 + 320t. a. What is the maximum height of the projectile? b. When does the projectile hit the ground?

This example solves a projectile motion problem using a quadratic equation: h(t) = -16t^2 + 320t. Learn how to calculate the maximum height (1600 feet) and the time when the projectile hits the ground (20 seconds). Perfect for Grades 9-12 students covering Algebra and Physics concepts.

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