A projectile is fired vertically, and its height (in feet) after t seconds is given by s(t)=−8t2+640t+2688s(t)=−8t2+640t+2688.

Step 3 of 3: 

At what rate is the projectile falling when it hits the ground? Round your final answer to the nearest hundredth.

To determine the rate at which the projectile is falling when it hits the ground, we need to follow these steps:

1. **Find the time when the projectile hits the ground**:
   The projectile hits the ground when \( s(t) = 0 \). So we need to solve the equation:
   \[
   s(t) = -8t^2 + 640t + 2688 = 0
   \]

2. **Find the derivative of the height function to get the velocity function**:
   The velocity function \( v(t) \) is the first derivative of the height function \( s(t) \).

3. **Evaluate the velocity at the time when the projectile hits the ground**:
   Use the time from step 1 in the velocity function to find the rate at which the projectile is falling.

### Step 1: Solve for \( t \) when \( s(t) = 0 \)

Given the equation:
\[
-8t^2 + 640t + 2688 = 0
\]

Divide the entire equation by -8 to simplify:
\[
t^2 - 80t - 336 = 0
\]

Solve this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -80 \), and \( c = -336 \):

\[
t = \frac{-(-80) \pm \sqrt{(-80)^2 - 4(1)(-336)}}{2(1)}
\]

\[
t = \frac{80 \pm \sqrt{6400 + 1344}}{2}
\]

\[
t = \frac{80 \pm \sqrt{7744}}{2}
\]

\[
t = \frac{80 \pm 88}{2}
\]

This gives us two solutions:
\[
t = \frac{80 + 88}{2} = \frac{168}{2} = 84
\]
\[
t = \frac{80 - 88}{2} = \frac{-8}{2} = -4
\]

Since time cannot be negative, the projectile hits the ground at \( t = 84 \) seconds.

### Step 2: Find the velocity function

The height function is:
\[
s(t) = -8t^2 + 640t + 2688
\]

Take the derivative to find the velocity function \( v(t) \):
\[
v(t) = \frac{d}{dt} \left( -8t^2 + 640t + 2688 \right)
\]

\[
v(t) = -16t + 640
\]

### Step 3: Evaluate the velocity at \( t = 84 \)

Substitute \( t = 84 \) into the velocity function:
\[
v(84) = -16(84) + 640
\]

\[
v(84) = -1344 + 640
\]

\[
v(84) = -704
\]

Therefore, the rate at which the projectile is falling when it hits the ground is \( -704 \) feet per second. Since the rate of falling is asked for, the negative sign indicates the direction (downwards), so the rate is:

\[
\boxed{704 \text{ feet per second}}
\]

Would you like more details on any step or further explanation?

Here are some related questions:
1. What is the significance of the negative sign in the velocity?
2. How can the quadratic formula be applied to other types of projectile motion problems?
3. What are other methods for solving quadratic equations?
4. How do you interpret the velocity function in terms of motion?
5. What is the physical meaning of the derivative of a position function?
6. Can you explain the difference between velocity and speed?
7. How does the initial height affect the time it takes for the projectile to hit the ground?
8. What are some real-world applications of these principles in physics?

**Tip:** When solving quadratic equations, always consider the context of the problem to determine the appropriate solution, such as ignoring negative time in this case.

Explore how to calculate the rate at which a projectile falls upon impact using a given height function and derivative. Step-by-step solution provided for a high school level problem in projectile motion and derivatives.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation