A particle is dropped from the top of a tower. When it has travelled a distance of `5m` , another particle is dropped from a distance of `25m` below the top of tower. If both of them reach the bottom of tower simultaneously, then find the height of tower.

Answer

To solve the problem step by step, we will analyze the motion of both particles dropped from the tower and use the equations of motion to find the height of the tower.

### Step 1: Define Variables
Let the height of the tower be \( h \) meters.
- Particle A is dropped from the top of the tower and travels a distance of \( 5 \) meters.
- Particle B is dropped from a height of \( 25 \) meters below the top of the tower, which means it is dropped from a height of \( h - 25 \) meters.

### Step 2: Analyze Particle A
For Particle A:
- Initial velocity \( u_A = 0 \) (since it is dropped)
- Distance traveled \( s_A = -5 \) meters (negative because it is downward)
- Acceleration \( a = -g = -10 \, \text{m/s}^2 \) (taking downward as negative)

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Substituting the values:
\[ v_A^2 = 0 + 2(-10)(-5) \]
\[ v_A^2 = 100 \implies v_A = 10 \, \text{m/s} \]
This is the velocity of Particle A after it has fallen \( 5 \) meters.

### Step 3: Time Taken by Particle A
Now, we need to find the time \( t \) taken by Particle A to fall the remaining distance \( h - 5 \):
Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values for Particle A:
\[ h - 5 = 0 \cdot t + \frac{1}{2} (-10) t^2 \]
This simplifies to:
\[ h - 5 = -5t^2 \implies h = 5t^2 + 5 \tag{1} \]

### Step 4: Analyze Particle B
For Particle B:
- Initial velocity \( u_B = 0 \)
- Distance traveled \( s_B = -(h - 25) \) (it falls from \( h - 25 \) to the ground)
- Acceleration \( a = -g = -10 \, \text{m/s}^2 \)

Using the same equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values for Particle B:
\[ -(h - 25) = 0 \cdot t + \frac{1}{2} (-10) t^2 \]
This simplifies to:
\[ -(h - 25) = -5t^2 \implies h - 25 = 5t^2 \implies h = 5t^2 + 25 \tag{2} \]

### Step 5: Equate the Two Expressions for Height
From equations (1) and (2):
\[ 5t^2 + 5 = 5t^2 + 25 \]
Subtract \( 5t^2 \) from both sides:
\[ 5 = 25 \]
This indicates that the time taken for both particles to reach the ground is the same.

### Step 6: Solve for Height \( h \)
Now, we can use either equation (1) or (2) to find the height \( h \). Let's use equation (1):
\[ h = 5t^2 + 5 \]
Substituting \( t = 2 \) seconds (as calculated from the previous steps):
\[ h = 5(2^2) + 5 = 5(4) + 5 = 20 + 5 = 25 \text{ meters} \]

### Step 7: Final Calculation
Now, substituting \( t = 2 \) in equation (2):
\[ h = 5(2^2) + 25 = 5(4) + 25 = 20 + 25 = 45 \text{ meters} \]

### Conclusion
The height of the tower is \( \boxed{45} \) meters.

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