A stone is dropped from the top of a tower of height `h`. Aftre `1 s` another stone is droppped from the balcony `20 m` below the top. Both reach the bottom simultaneously. What is the value of `h` ? Take `g=10 ms^(-2)`.

Options

A. `31 5m`
B. ` 312.5 m`
C. ` 31.25 m`
D. ` 25,31`

Explanation

To solve the problem step by step, we can follow the outlined reasoning and calculations:

### Step 1: Understand the Problem
A stone is dropped from the top of a tower of height \( h \). After 1 second, another stone is dropped from a point 20 meters below the top of the tower. Both stones reach the ground simultaneously. We need to find the height \( h \) of the tower.

### Step 2: Analyze the First Stone
For the first stone dropped from height \( h \):
- Initial velocity \( u = 0 \) (since it is dropped)
- Distance \( s = h \)
- Time taken to reach the ground = \( t \)
- Acceleration \( a = g = 10 \, \text{m/s}^2 \)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ h = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ h = 5t^2 \quad \text{(Equation 1)} \]

### Step 3: Analyze the Second Stone
The second stone is dropped from a height of \( h - 20 \) meters (since it is dropped from 20 meters below the top):
- Initial velocity \( u = 0 \)
- Distance \( s = h - 20 \)
- Time taken to reach the ground = \( t - 1 \) seconds (since it is dropped 1 second later)
- Acceleration \( a = g = 10 \, \text{m/s}^2 \)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ h - 20 = 0 \cdot (t - 1) + \frac{1}{2} \cdot 10 \cdot (t - 1)^2 \]
This simplifies to:
\[ h - 20 = 5(t - 1)^2 \quad \text{(Equation 2)} \]

### Step 4: Expand Equation 2
Expanding \( (t - 1)^2 \):
\[ (t - 1)^2 = t^2 - 2t + 1 \]
So, substituting back:
\[ h - 20 = 5(t^2 - 2t + 1) \]
This simplifies to:
\[ h - 20 = 5t^2 - 10t + 5 \]
Rearranging gives:
\[ h = 5t^2 - 10t + 25 \quad \text{(Equation 3)} \]

### Step 5: Set Equations 1 and 3 Equal
From Equation 1, we have:
\[ h = 5t^2 \]
From Equation 3, we have:
\[ h = 5t^2 - 10t + 25 \]
Setting these equal to each other:
\[ 5t^2 = 5t^2 - 10t + 25 \]
This simplifies to:
\[ 0 = -10t + 25 \]
Rearranging gives:
\[ 10t = 25 \implies t = 2.5 \, \text{s} \]

### Step 6: Substitute \( t \) Back to Find \( h \)
Now substitute \( t = 2.5 \) seconds back into Equation 1:
\[ h = 5t^2 = 5(2.5)^2 = 5 \cdot 6.25 = 31.25 \, \text{m} \]

### Final Answer
The height of the tower \( h \) is \( 31.25 \, \text{m} \).

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