A stone is relased from top of a lower. It covers a distance of 80m in last 2 seconds of its motion. Then the height of the lower is : `(g = 10 m//s^(2))`

Options

A. `320m`
B. `245m`
C. `180m`
D. `125m`

Explanation

To solve the problem step by step, we need to find the height of the tower from which the stone is released. We know that the stone covers a distance of 80 meters in the last 2 seconds of its motion.

### Step 1: Define the Variables
Let:
- \( h \) = height of the tower
- \( N \) = total time taken by the stone to reach the ground (in seconds)
- \( g \) = acceleration due to gravity = \( 10 \, \text{m/s}^2 \)

### Step 2: Use the Distance Formula
The distance covered by the stone in the last 2 seconds can be expressed as:
\[ S_N - S_{N-2} = 80 \, \text{m} \]
Where:
- \( S_N \) = total distance covered in \( N \) seconds
- \( S_{N-2} \) = distance covered in \( N-2 \) seconds

### Step 3: Apply the Equation of Motion
Using the equation of motion:
\[ S = ut + \frac{1}{2} g t^2 \]
Since the stone is released from rest, \( u = 0 \):
\[ S_N = \frac{1}{2} g N^2 \]
\[ S_{N-2} = \frac{1}{2} g (N-2)^2 \]

### Step 4: Substitute the Values
Substituting these into the equation gives:
\[ \frac{1}{2} g N^2 - \frac{1}{2} g (N-2)^2 = 80 \]
Substituting \( g = 10 \):
\[ \frac{1}{2} \cdot 10 N^2 - \frac{1}{2} \cdot 10 (N-2)^2 = 80 \]
This simplifies to:
\[ 5 N^2 - 5 (N^2 - 4N + 4) = 80 \]
\[ 5 N^2 - 5 N^2 + 20N - 20 = 80 \]
\[ 20N - 20 = 80 \]
\[ 20N = 100 \]
\[ N = 5 \, \text{seconds} \]

### Step 5: Calculate the Height of the Tower
Now, we can find the height of the tower using the total time:
\[ h = S_N = \frac{1}{2} g N^2 \]
Substituting \( N = 5 \):
\[ h = \frac{1}{2} \cdot 10 \cdot (5^2) = \frac{1}{2} \cdot 10 \cdot 25 = 125 \, \text{meters} \]

### Final Answer
The height of the tower is \( \boxed{125 \, \text{meters}} \).