From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is doubleof what it was at a height h above A. Show that the greatest height attained by the stone is `5/3 h.`

Explanation

To solve the problem, we need to analyze the motion of the stone projected vertically upwards from point A and apply the equations of motion.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- Let the initial velocity of the stone when projected from point A be \( U \).
- When the stone reaches a height \( h \) above point A, let its velocity be \( V \).
- When the stone descends to a height \( h \) below point A, its velocity is given as \( 2V \).

2. **Applying the First Condition (At height \( h \) above A):**
- Using the equation of motion:
\[ V^2 = U^2 - 2gh \]
- This equation relates the initial velocity \( U \), final velocity \( V \), and the height \( h \) above point A.

3. **Applying the Second Condition (At height \( h \) below A):**
- When the stone is \( h \) below point A, its velocity is \( 2V \). The height from point A is now \( -h \).
- Using the equation of motion again:
\[ (2V)^2 = U^2 - 2g(-h) \]
- Simplifying this gives:
\[ 4V^2 = U^2 + 2gh \]

4. **Setting Up the Equations:**
- From the first equation, we have:
\[ V^2 = U^2 - 2gh \quad \text{(1)} \]
- From the second equation, we have:
\[ 4V^2 = U^2 + 2gh \quad \text{(2)} \]

5. **Substituting Equation (1) into Equation (2):**
- Substitute \( V^2 \) from equation (1) into equation (2):
\[ 4(U^2 - 2gh) = U^2 + 2gh \]
- Expanding and rearranging gives:
\[ 4U^2 - 8gh = U^2 + 2gh \]
\[ 3U^2 = 10gh \]
- Therefore, we can express \( U^2 \) as:
\[ U^2 = \frac{10}{3}gh \quad \text{(3)} \]

6. **Finding the Maximum Height:**
- The maximum height \( H \) attained by the stone can be calculated using the formula:
\[ H = \frac{U^2}{2g} \]
- Substituting \( U^2 \) from equation (3):
\[ H = \frac{\frac{10}{3}gh}{2g} = \frac{10h}{6} = \frac{5h}{3} \]

7. **Conclusion:**
- Thus, the greatest height attained by the stone is:
\[ H = \frac{5}{3}h \]
- This confirms the statement in the problem.