A rock is tossed straight up from the ground with a speed of 21 m/s. When it returns, it falls into a hole 10 m deep. What is the rock's speed as it hits the bottom of the hole?

We can solve this problem using the principle of energy conservation, assuming no air resistance. The total mechanical energy of the system remains constant, so we can use the initial kinetic energy and gravitational potential energy to find the final speed.

### Given:
- Initial speed \( v_0 = 21 \, \text{m/s} \)
- Depth of the hole \( h = 10 \, \text{m} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

### Solution:

1. **Initial energy (at the ground level):**
   The rock is initially tossed upward, so its total energy is purely kinetic:
   \[
   E_{\text{initial}} = \frac{1}{2} m v_0^2
   \]
   where \( v_0 \) is the initial speed and \( m \) is the mass of the rock (which cancels out later).

2. **Final energy (when the rock hits the bottom of the hole):**
   At the bottom of the hole, the rock has both kinetic and potential energy. The potential energy relative to the ground is negative due to the hole being 10 m deep:
   \[
   E_{\text{final}} = \frac{1}{2} m v^2 + mgh
   \]
   where \( v \) is the final speed, \( h = -10 \, \text{m} \), and \( g = 9.8 \, \text{m/s}^2 \).

3. **Conservation of energy:**
   Since energy is conserved, we equate the initial energy to the final energy:
   \[
   \frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + mgh
   \]
   Simplifying (the mass \( m \) cancels out):
   \[
   \frac{1}{2} v_0^2 = \frac{1}{2} v^2 + gh
   \]
   \[
   \frac{1}{2} v^2 = \frac{1}{2} v_0^2 - gh
   \]
   \[
   v^2 = v_0^2 - 2gh
   \]
   Substituting the values:
   \[
   v^2 = 21^2 - 2(9.8)(-10)
   \]
   \[
   v^2 = 441 + 196
   \]
   \[
   v^2 = 637
   \]
   \[
   v = \sqrt{637} \approx 25.24 \, \text{m/s}
   \]

Thus, the rock's speed as it hits the bottom of the hole is approximately **25.24 m/s**.

Would you like further details or have any questions? 

Here are some related questions for further exploration:
1. How would the speed change if the hole were 20 m deep?
2. What is the maximum height the rock reaches before falling?
3. How much time does the rock take to reach the bottom of the hole?
4. How would air resistance affect the final speed?
5. What is the total distance traveled by the rock from launch to the bottom of the hole?

**Tip:** Energy conservation is a powerful tool to solve problems where forces like gravity are involved, especially when the path or time isn't directly needed.

Solve the physics problem of a rock tossed upward with an initial speed of 21 m/s and falling into a 10 m deep hole. Learn how to apply the conservation of energy principle to determine the rock's speed as it hits the bottom. Suitable for high school students in Grades 10-12.

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