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^@)`

Options

A. (a) `10 m`
B. (b) ` 45 m`
C. (c ) ` 80`
D. (d)` 50 m`

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration due to gravity.

### Step 1: Understand the problem
We have a body dropped from the top of a tower, and it falls a distance of 40 meters during the last 2 seconds of its fall. We need to find the height of the tower.

### Step 2: Define the variables
- Let \( H \) be the height of the tower.
- Let \( T \) be the total time taken for the body to reach the ground.
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

### Step 3: Use the equation of motion
The distance fallen by the body in \( T \) seconds can be calculated using the equation:
\[ H = \frac{1}{2} g T^2 \]
Substituting the value of \( g \):
\[ H = \frac{1}{2} \times 10 \times T^2 = 5T^2 \quad \text{(Equation 1)} \]

### Step 4: Calculate the distance fallen in the last 2 seconds
The distance fallen in the last 2 seconds can be found using the equation:
\[ S = \frac{1}{2} g (T^2) - \frac{1}{2} g ((T-2)^2) \]
This gives us the distance fallen in the last 2 seconds:
\[ S = 5T^2 - 5(T-2)^2 \]

### Step 5: Simplify the expression for \( S \)
Expanding \( (T-2)^2 \):
\[ (T-2)^2 = T^2 - 4T + 4 \]
Now substituting back:
\[ S = 5T^2 - 5(T^2 - 4T + 4) = 5T^2 - 5T^2 + 20T - 20 = 20T - 20 \]

### Step 6: Set the distance fallen in the last 2 seconds equal to 40 m
According to the problem, the distance fallen in the last 2 seconds is 40 m:
\[ 20T - 20 = 40 \]
Solving for \( T \):
\[ 20T = 60 \implies T = 3 \, \text{s} \]

### Step 7: Substitute \( T \) back into the equation for height
Now, substitute \( T = 3 \) seconds back into Equation 1 to find \( H \):
\[ H = 5T^2 = 5 \times (3^2) = 5 \times 9 = 45 \, \text{m} \]

### Final Answer
The height of the tower is \( H = 45 \, \text{m} \).

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