https://file.aiquickdraw.com/mathbot/file/17284387899825q4uy3rq.jpg

Let's go through the questions one by one for problem 27. The position function for free-falling objects is given as:

\[
s(t) = -16t^2 + v_0t + s_0
\]

Where:
- \( v_0 \) is the initial velocity
- \( s_0 \) is the initial position (height of the object)
- \( t \) is the time in seconds
- \( s(t) \) is the height at time \( t \)
- The constant \(-16\) comes from half the acceleration due to gravity in feet per second squared (since \( g = 32 \, \text{ft/s}^2 \)).

Given:
- Initial velocity \( v_0 = -30 \, \text{ft/s} \)
- Initial position \( s_0 = 600 \, \text{ft} \)

### (a) Position and Velocity Functions
The position function is:

\[
s(t) = -16t^2 + (-30)t + 600 = -16t^2 - 30t + 600
\]

The velocity function is the derivative of the position function with respect to time \( t \):

\[
v(t) = \frac{d}{dt}\left( -16t^2 - 30t + 600 \right) = -32t - 30
\]

### (b) Average Velocity on the Interval \([1, 3]\)

The average velocity is calculated by:

\[
\text{Average velocity} = \frac{s(3) - s(1)}{3 - 1}
\]

Let's find \( s(3) \) and \( s(1) \):

- \( s(3) = -16(3)^2 - 30(3) + 600 = -144 - 90 + 600 = 366 \, \text{ft} \)
- \( s(1) = -16(1)^2 - 30(1) + 600 = -16 - 30 + 600 = 554 \, \text{ft} \)

Now calculate the average velocity:

\[
\text{Average velocity} = \frac{366 - 554}{3 - 1} = \frac{-188}{2} = -94 \, \text{ft/s}
\]

### (c) Instantaneous Velocities at \( t = 1 \) and \( t = 3 \)

Using the velocity function \( v(t) = -32t - 30 \):

- At \( t = 1 \): \( v(1) = -32(1) - 30 = -32 - 30 = -62 \, \text{ft/s} \)
- At \( t = 3 \): \( v(3) = -32(3) - 30 = -96 - 30 = -126 \, \text{ft/s} \)

### (d) Time to Reach Ground Level

To find the time when the ball hits the ground, we set \( s(t) = 0 \):

\[
-16t^2 - 30t + 600 = 0
\]

Solving this quadratic equation using the quadratic formula:

\[
t = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(-16)(600)}}{2(-16)}
\]
\[
t = \frac{30 \pm \sqrt{900 + 38400}}{-32} = \frac{30 \pm \sqrt{39300}}{-32}
\]

Approximating the square root:

\[
t = \frac{30 \pm 198.24}{-32}
\]

Only the positive root is physically meaningful:

\[
t = \frac{30 - 198.24}{-32} = \frac{-168.24}{-32} \approx 5.26 \, \text{seconds}
\]

### (e) Velocity at Impact

At impact, we use \( t \approx 5.26 \) in the velocity function:

\[
v(5.26) = -32(5.26) - 30 \approx -168.32 - 30 = -198.32 \, \text{ft/s}
\]

The velocity at impact is approximately \( -198.32 \, \text{ft/s} \).

---

Do you want more details on any specific part of this? 

Here are five additional questions to consider:
1. What is the significance of the negative sign in the velocity values?
2. How would the velocity function change if the ball were thrown upward instead of downward?
3. Could you determine the maximum height reached if the ball had been thrown upward?
4. How would the problem change if the initial velocity or height were different?
5. What physical factors are neglected in this model of free-fall?

**Tip:** When solving quadratic equations for time in free-fall problems, always check which root is physically meaningful, as time cannot be negative.

A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. Determine the position and velocity functions, average velocity, instantaneous velocity, time to reach the ground, and the velocity at impact.

This problem explores the kinematic motion of a ball thrown from a 600-foot building with an initial velocity of -30 feet per second. Learn how to calculate the position and velocity functions, find average and instantaneous velocities, determine the time to hit the ground, and the impact velocity. This is suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation