A body dropped from top of a tower falls through ` 40 m` during the last two seconds of its fall. The height of tower in m is ( g= 10 m//s^2)`

Options

A. `60m`
B. `45m`
C. `80m`
D. `50m`

Answer

To find the height of the tower from which a body is dropped, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is dropped from the top of a tower and falls through a distance of 40 m during the last 2 seconds of its fall.
- We need to find the height of the tower (h) given that the acceleration due to gravity (g) is 10 m/s².

2. **Using the Equation of Motion**:
- The distance fallen by the body can be described using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance fallen, \( u \) is the initial velocity (which is 0 since the body is dropped), \( a \) is the acceleration due to gravity (g), and \( t \) is the total time of fall.

3. **Setting Up the Equation for Total Fall**:
- Since the initial velocity \( u = 0 \), the equation simplifies to:
\[ h = \frac{1}{2} g t^2 \]
- This will be our Equation (1).

4. **Distance Fallen in the Last 2 Seconds**:
- The distance fallen in the last 2 seconds can be expressed as:
\[ s_{last\ 2\ seconds} = s(t) - s(t-2) \]
- We know this distance is 40 m. Therefore:
\[ s(t) - s(t-2) = 40 \]

5. **Calculating Distances**:
- Using the equation of motion for both \( s(t) \) and \( s(t-2) \):
\[ s(t) = \frac{1}{2} g t^2 \]
\[ s(t-2) = \frac{1}{2} g (t-2)^2 \]
- Substitute these into the distance equation:
\[ \frac{1}{2} g t^2 - \frac{1}{2} g (t-2)^2 = 40 \]
- Factoring out \( \frac{1}{2} g \):
\[ \frac{1}{2} g \left( t^2 - (t^2 - 4t + 4) \right) = 40 \]
- Simplifying gives:
\[ \frac{1}{2} g (4t - 4) = 40 \]
- Rearranging leads to:
\[ 2g(t - 1) = 40 \]

6. **Solving for Time (t)**:
- Substitute \( g = 10 \):
\[ 20(t - 1) = 40 \]
- Dividing both sides by 20:
\[ t - 1 = 2 \implies t = 3 \text{ seconds} \]

7. **Finding the Height of the Tower**:
- Now substitute \( t = 3 \) back into Equation (1):
\[ h = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45 \text{ m} \]

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

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