A ball is thrown straight up so that it reaches a height of 25m. How fast was it going when it was 5m high?

To solve this problem, we can use the principles of energy conservation, specifically the conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) of the ball remains constant throughout its motion, assuming there is no air resistance.

### Given:
- Maximum height, \( h_{\text{max}} = 25 \) m
- Height when speed is to be determined, \( h = 5 \) m

### Steps:

1. **Determine the initial speed at 5m:**

   When the ball is at height \( h = 5 \) m, its total mechanical energy is the sum of its kinetic energy and potential energy.

   - **Potential Energy at height \( h \):**
     \[
     PE = mgh
     \]
     where:
     - \( m \) is the mass of the ball (we'll see that it cancels out),
     - \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
     - \( h = 5 \, \text{m} \).
  
   - **Kinetic Energy at height \( h \):**
     \[
     KE = \frac{1}{2}mv^2
     \]
     where \( v \) is the velocity at height \( h \).

   At the maximum height \( h_{\text{max}} = 25 \) m, all the kinetic energy has been converted into potential energy, so:
     \[
     mgh_{\text{max}} = \frac{1}{2}mv_{\text{initial}}^2
     \]
     Here, \( v_{\text{initial}} \) is the initial speed of the ball at ground level.

2. **Use the energy conservation principle between \( h_{\text{max}} \) and \( h \):**

   The total mechanical energy at any point during the ball's flight is the same as it was initially. Therefore:
     \[
     mgh + \frac{1}{2}mv^2 = mgh_{\text{max}}
     \]
   Simplifying (mass \( m \) cancels out):
     \[
     gh + \frac{1}{2}v^2 = gh_{\text{max}}
     \]
     \[
     \frac{1}{2}v^2 = g(h_{\text{max}} - h)
     \]
     \[
     v^2 = 2g(h_{\text{max}} - h)
     \]
     \[
     v = \sqrt{2g(h_{\text{max}} - h)}
     \]
  
3. **Substitute the values:**
   \[
   v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times (25 \, \text{m} - 5 \, \text{m})}
   \]
   \[
   v = \sqrt{2 \times 9.8 \times 20}
   \]
   \[
   v = \sqrt{392} \, \text{m/s}
   \]
   \[
   v \approx 19.8 \, \text{m/s}
   \]

### Conclusion:
The ball was going approximately **19.8 m/s** when it was at a height of 5 meters.

Do you want more details on this or have any questions?

Here are some related questions to consider:
1. How would air resistance affect the ball's speed at 5 meters?
2. What was the initial velocity of the ball when it was thrown?
3. How much time does it take for the ball to reach the maximum height?
4. What is the velocity of the ball when it returns to the 5-meter height on its way down?
5. How does the mass of the ball affect the speed calculation?

**Tip:** When dealing with energy conservation, remember that potential energy is maximum at the peak height, and kinetic energy is maximum at the initial and final points if we neglect air resistance.

Learn how to calculate the velocity of a ball thrown straight up to reach a height of 25 meters. Understand the principles of energy conservation in physics, including potential and kinetic energy equations. Suitable for high school physics students.

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