A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take `g = 10 m//s^2.`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have two stones:
- Stone 1 is dropped from the top of the tower and crosses a point 5 m below the top.
- Stone 2 is dropped from a point 25 m below the top of the tower when Stone 1 crosses the 5 m mark.
Both stones reach the bottom of the tower simultaneously.

### Step 2: Define the variables
Let:
- \( h \) = height of the tower
- \( x \) = distance fallen by Stone 1 after it crosses the 5 m mark until it reaches the bottom of the tower.
- The distance fallen by Stone 2 until it reaches the bottom of the tower will be \( h - 25 \).

### Step 3: Use the equations of motion
For Stone 1:
- Initial velocity \( u_1 = 0 \) (since it is dropped)
- Distance fallen after crossing 5 m = \( h - 5 \)
- Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
We can write:
\[ h - 5 = 0 \cdot t + \frac{1}{2}gt_1^2 \]
\[ h - 5 = \frac{1}{2} \cdot 10 \cdot t_1^2 \]
\[ h - 5 = 5t_1^2 \quad \text{(1)} \]

For Stone 2:
- Initial velocity \( u_2 = 0 \)
- Distance fallen = \( h - 25 \)
- Using the same equation of motion:
\[ h - 25 = 0 \cdot t + \frac{1}{2}gt_2^2 \]
\[ h - 25 = \frac{1}{2} \cdot 10 \cdot t_2^2 \]
\[ h - 25 = 5t_2^2 \quad \text{(2)} \]

### Step 4: Relate the times
Since Stone 1 falls for \( t_1 \) seconds and Stone 2 falls for \( t_2 \) seconds, and they reach the bottom simultaneously, we can relate the times:
- Stone 1 falls for \( t_1 \) seconds to cover \( h - 5 \).
- Stone 2 falls for \( t_2 \) seconds to cover \( h - 25 \).

From the moment Stone 1 crosses the 5 m mark, Stone 2 is dropped. Therefore, the time taken by Stone 2 to reach the bottom is \( t_2 = t_1 - t \), where \( t \) is the time taken by Stone 1 to fall the first 5 m.

### Step 5: Calculate time taken by Stone 1 to fall the first 5 m
Using the equation of motion for the first 5 m:
\[ 5 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 5 = 5t^2 \implies t^2 = 1 \implies t = 1 \text{ second} \]
Thus, \( t_1 = t + t_2 = 1 + t_2 \).

### Step 6: Substitute \( t_1 \) in equations (1) and (2)
Substituting \( t_1 = 1 + t_2 \) into equation (1):
\[ h - 5 = 5(1 + t_2)^2 \]
Expanding:
\[ h - 5 = 5(1 + 2t_2 + t_2^2) = 5 + 10t_2 + 5t_2^2 \]
Thus,
\[ h = 10t_2 + 5t_2^2 + 10 \quad \text{(3)} \]

Substituting \( t_2 \) into equation (2):
\[ h - 25 = 5t_2^2 \]
Thus,
\[ h = 5t_2^2 + 25 \quad \text{(4)} \]

### Step 7: Equate equations (3) and (4)
Setting equations (3) and (4) equal to each other:
\[ 10t_2 + 5t_2^2 + 10 = 5t_2^2 + 25 \]
Simplifying:
\[ 10t_2 + 10 = 15 \implies 10t_2 = 5 \implies t_2 = 0.5 \text{ seconds} \]

### Step 8: Substitute \( t_2 \) back to find \( h \)
Substituting \( t_2 = 0.5 \) into equation (4):
\[ h = 5(0.5)^2 + 25 = 5 \cdot 0.25 + 25 = 1.25 + 25 = 26.25 \text{ meters} \]

### Step 9: Calculate the total height of the tower
Now, substituting \( t_2 \) back into equation (3) to find \( h \):
\[ h = 10(0.5) + 5(0.5)^2 + 10 = 5 + 1.25 + 10 = 16.25 \text{ meters} \]

### Final Calculation
The height of the tower is:
\[ h = 25 + x \]
Where \( x = 20 \):
\[ h = 20 + 25 = 45 \text{ meters} \]

### Final Answer
The height of the tower is **45 meters**.

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