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 step by step, we will use the equations of motion under uniform acceleration due to gravity. Let's denote:

- \( u \): Initial velocity of the stone when projected upwards.
- \( v_h \): Velocity of the stone when it is at a height \( h \) above point A.
- \( v_{-h} \): Velocity of the stone when it is at a distance \( h \) below point A.
- \( g \): Acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- \( h_{max} \): Maximum height attained by the stone.

### Step 1: Establish the relationship between velocities
According to the problem, when the stone is at a distance \( h \) below point A, its velocity \( v_{-h} \) is double the velocity \( v_h \) at a height \( h \) above A. Thus, we can write:

\[ v_{-h} = 2v_h \]

### Step 2: Apply the equations of motion
Using the equation of motion for the stone at height \( h \) above point A:

\[ v_h^2 = u^2 - 2gh \]

And for the stone at a distance \( h \) below point A:

\[ v_{-h}^2 = u^2 + 2gh \]

### Step 3: Substitute the relationship between velocities
From our earlier relationship, we can substitute \( v_{-h} \) in terms of \( v_h \):

\[ (2v_h)^2 = u^2 + 2gh \]

This simplifies to:

\[ 4v_h^2 = u^2 + 2gh \]

### Step 4: Substitute \( v_h^2 \) from the first equation
Now we can substitute \( v_h^2 \) from the first equation into this equation:

\[ 4(u^2 - 2gh) = u^2 + 2gh \]

### Step 5: Expand and simplify the equation
Expanding the left side gives:

\[ 4u^2 - 8gh = u^2 + 2gh \]

Now, rearranging the equation:

\[ 4u^2 - u^2 = 8gh + 2gh \]

This leads to:

\[ 3u^2 = 10gh \]

### Step 6: Solve for \( u^2 \)
Dividing both sides by 3 gives:

\[ u^2 = \frac{10gh}{3} \]

### Step 7: Calculate the maximum height
The maximum height \( h_{max} \) attained by the stone can be calculated using the formula:

\[ h_{max} = \frac{u^2}{2g} \]

Substituting \( u^2 \):

\[ h_{max} = \frac{\frac{10gh}{3}}{2g} = \frac{10h}{6} = \frac{5h}{3} \]

### Conclusion
Thus, we have shown that the greatest height attained by the stone is:

\[ h_{max} = \frac{5h}{3} \]