https://file.aiquickdraw.com/mathbot/file/17304424963642jokkn5s.jpg

Let's go through each part of the question step-by-step.

Given:
- Initial height \( h_0 = 650 \) feet
- Initial velocity \( v_0 = 20 \) feet per second (upwards)
- Acceleration due to gravity \( g = -32 \) feet per second squared

### Part A: Height of the stone 4 seconds later

The height \( h(t) \) of the stone at any time \( t \) can be found using the equation:
\[
h(t) = h_0 + v_0 t + \frac{1}{2} g t^2
\]

Substitute the values:
\[
h(4) = 650 + 20 \times 4 + \frac{1}{2} \times (-32) \times 4^2
\]

Breaking it down:
1. \( v_0 t = 20 \times 4 = 80 \)
2. \( \frac{1}{2} \times (-32) \times 4^2 = -16 \times 16 = -256 \)

So:
\[
h(4) = 650 + 80 - 256 = 474 \text{ feet}
\]

### Part B: Time when the stone hits the ground

The stone will hit the ground when \( h(t) = 0 \). Using the height equation:
\[
0 = 650 + 20t - 16t^2
\]

Rearrange it to form a standard quadratic equation:
\[
-16t^2 + 20t + 650 = 0
\]

To solve for \( t \), we can use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -16 \), \( b = 20 \), and \( c = 650 \).

Calculating:
1. \( b^2 - 4ac = 20^2 - 4 \times (-16) \times 650 \)
2. \( = 400 + 41600 = 42000 \)
3. \( \sqrt{42000} \approx 204.94 \)

So:
\[
t = \frac{-20 \pm 204.94}{-32}
\]

This yields two values for \( t \):
- \( t_1 = \frac{-20 + 204.94}{-32} \approx 7.22 \) seconds (positive, so this is the solution)
- \( t_2 = \frac{-20 - 204.94}{-32} \) (negative, discard)

Thus, the stone hits the ground at approximately \( t = 7.22 \) seconds.

### Part C: Velocity of the stone when it hits the ground

The velocity \( v(t) \) at any time \( t \) is given by:
\[
v(t) = v_0 + g \cdot t
\]

Substitute \( t = 7.22 \) seconds:
\[
v(7.22) = 20 + (-32) \times 7.22
\]
\[
= 20 - 231.04 = -211.04 \text{ feet per second}
\]

The negative sign indicates the stone is moving downward at the time it hits the ground.

---

Would you like more details on any of the steps?

---

Here are some additional related questions:
1. What would be the maximum height reached by the stone?
2. How long does it take for the stone to reach its maximum height?
3. What would happen to the stone's final velocity if it were thrown with a higher initial velocity?
4. How would the height change if the stone were thrown from a higher building?
5. How would air resistance affect the time taken for the stone to hit the ground?

**Tip:** When solving projectile motion problems, be mindful of the direction of acceleration due to gravity, as it affects the sign in your equations.

A stone is thrown straight up from the edge of a roof, 650 feet above the ground, at a speed of 20 feet per second. A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 4 seconds later? B. At what time does the stone hit the ground? C. What is the velocity of the stone when it hits the ground?

Solve a projectile motion problem where a stone is thrown from a 650-feet-high roof at 20 ft/s upwards. Learn how to calculate the height of the stone after 4 seconds, the time it takes to hit the ground, and its final velocity. This problem covers key kinematic concepts and involves using the height and velocity equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation