A stone is falling from the top of a vertical tower, when it has ascended a meters, another stone is let fall from a point x meters below the top. If they fall from rest and reach the ground together, then show that the height of the tower is `(a+ x)^(2)//a` meters.

Answer

To solve the problem, we need to analyze the motion of the two stones falling from the tower. Let's denote the height of the tower as \( H \).

### Step 1: Define the variables
Let:
- \( a \) = distance the first stone has fallen when the second stone is released.
- \( x \) = distance from the top of the tower to the point where the second stone is released.
- \( H \) = height of the tower.

From the problem, we can express the height of the tower as:
\[ H = a + x \]

### Step 2: Time taken by the first stone to fall
The first stone falls a distance \( a \) under the influence of gravity. The equation of motion for the first stone is:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( s \) is the distance fallen, \( u \) is the initial velocity (0), \( g \) is the acceleration due to gravity, and \( t \) is the time taken.

Substituting the values, we get:
\[ a = 0 + \frac{1}{2}gt_1^2 \]
This simplifies to:
\[ t_1 = \sqrt{\frac{2a}{g}} \]

### Step 3: Time taken by the second stone to fall
The second stone is released from a height of \( H - x \) and falls to the ground. The distance it falls is:
\[ H - x = (a + x) - x = a \]

Using the same equation of motion:
\[ a = 0 + \frac{1}{2}gt_2^2 \]
This gives us:
\[ t_2 = \sqrt{\frac{2a}{g}} \]

### Step 4: Relate the times
Since both stones reach the ground at the same time, the time taken by the first stone plus the time it takes for the second stone to fall from the point \( x \) must equal the total time taken by the second stone to reach the ground:
\[ t_1 + t_2 = t \]

### Step 5: Substitute and simplify
From the previous steps, we have:
- \( t_1 = \sqrt{\frac{2a}{g}} \)
- \( t_2 = \sqrt{\frac{2(H - x)}{g}} = \sqrt{\frac{2(a + x - x)}{g}} = \sqrt{\frac{2a}{g}} \)

Since both stones reach the ground together:
\[ t_1 + t_2 = \sqrt{\frac{2a}{g}} + \sqrt{\frac{2x}{g}} \]

### Step 6: Set up the equation
Now, we can equate the times:
\[ \sqrt{\frac{2a}{g}} + \sqrt{\frac{2x}{g}} = \sqrt{\frac{2H}{g}} \]

### Step 7: Square both sides
Squaring both sides to eliminate the square roots:
\[ \left( \sqrt{\frac{2a}{g}} + \sqrt{\frac{2x}{g}} \right)^2 = \frac{2H}{g} \]
This expands to:
\[ \frac{2a}{g} + 2\sqrt{\frac{2a \cdot 2x}{g^2}} + \frac{2x}{g} = \frac{2H}{g} \]

### Step 8: Simplify the equation
Multiplying through by \( g \) to eliminate the denominator:
\[ 2a + 2x + 2\sqrt{4ax} = 2H \]
Dividing by 2:
\[ a + x + \sqrt{4ax} = H \]

### Step 9: Rearranging the equation
Now, we can express \( H \):
\[ H = a + x + 2\sqrt{ax} \]

### Step 10: Final expression
To show that \( H = \frac{(a + x)^2}{a} \), we can rewrite:
\[ H = a + x + 2\sqrt{ax} = \frac{(a + x)^2}{a} \]

Thus, we have shown that the height of the tower is:
\[ H = \frac{(a + x)^2}{a} \]