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. When the ball is at its maximum height, its kinetic energy has been completely converted to potential energy. As the ball moves down, potential energy converts back to kinetic energy.

### Given:
- Maximum height, \( h_{\text{max}} = 25 \) meters
- Height at the given point, \( h = 5 \) meters
- Acceleration due to gravity, \( g = 9.8 \ \text{m/s}^2 \)

### Approach:
1. **Initial Energy at Maximum Height (25 m):**
   - At 25 m, the velocity is 0 m/s.
   - The total mechanical energy is purely potential:  
     \[ E_{\text{max}} = m \cdot g \cdot h_{\text{max}} \]
   
2. **Energy at Height 5 m:**
   - At 5 m, the total mechanical energy is the sum of potential and kinetic energy:  
     \[ E_{\text{5m}} = m \cdot g \cdot h + \frac{1}{2} m v^2 \]
   
3. **Conservation of Mechanical Energy:**
   - The total energy at the maximum height should equal the total energy at any other point:  
     \[ m \cdot g \cdot h_{\text{max}} = m \cdot g \cdot h + \frac{1}{2} m v^2 \]
   - The mass \( m \) cancels out, leaving:  
     \[ g \cdot h_{\text{max}} = g \cdot h + \frac{1}{2} v^2 \]

4. **Solve for \( v \):**
   \[ v^2 = 2g \cdot (h_{\text{max}} - h) \]
   \[ v = \sqrt{2g \cdot (h_{\text{max}} - h)} \]

### Calculation:
\[ v = \sqrt{2 \cdot 9.8 \cdot (25 - 5)} \]
\[ v = \sqrt{2 \cdot 9.8 \cdot 20} \]
\[ v = \sqrt{392} \]
\[ v \approx 19.8 \ \text{m/s} \]

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

Would you like more details or have any further questions?

### Related Questions:
1. What is the ball's initial velocity when it was first thrown?
2. How long does it take for the ball to reach the maximum height?
3. What is the total time of flight for the ball?
4. How fast was the ball moving when it was 10 meters high?
5. What is the maximum height if the ball's initial speed was increased by 20%?

### Tip:
In problems involving free fall or projectile motion, conservation of energy is often a simpler method than kinematic equations for finding velocities at different heights.

Learn how to calculate the initial velocity of a ball thrown upwards using energy conservation principles. This problem involves determining the velocity when the ball reaches a height of 5 meters after reaching a maximum height of 25 meters. Detailed step-by-step solution included.

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