From a tower of height 45m, a stone is dropped, calculate the total distance travelled by the stone during the last one second of motion? (Given: `g = 10 m//s^(2)`)

Options

A. 4.5 m
B. 10 m
C. 25 m
D. 35 m

Explanation

To solve the problem of calculating the total distance traveled by the stone during the last one second of its motion from a height of 45 meters, we can follow these steps:

### Step 1: Determine the time taken to fall from the tower
We start with the equation of motion for an object under free fall:
\[ H = U t + \frac{1}{2} g t^2 \]
Where:
- \( H = 45 \, \text{m} \) (height of the tower)
- \( U = 0 \, \text{m/s} \) (initial velocity, since the stone is dropped)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values into the equation:
\[ 45 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 45 = 5t^2 \]
Now, solving for \( t^2 \):
\[ t^2 = \frac{45}{5} = 9 \]
Taking the square root gives:
\[ t = 3 \, \text{s} \]

### Step 2: Calculate the distance fallen in the first 2 seconds
Next, we need to find out how far the stone has fallen in the first 2 seconds. We use the same equation of motion:
\[ H_2 = U t + \frac{1}{2} g t^2 \]
For \( t = 2 \, \text{s} \):
\[ H_2 = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 \]
This simplifies to:
\[ H_2 = 5 \cdot 4 = 20 \, \text{m} \]

### Step 3: Calculate the distance traveled in the last second
To find the distance traveled during the last second (from 2 seconds to 3 seconds), we subtract the distance fallen in the first 2 seconds from the total height:
\[ \text{Distance in last second} = H - H_2 \]
\[ \text{Distance in last second} = 45 \, \text{m} - 20 \, \text{m} = 25 \, \text{m} \]

### Final Answer
The total distance traveled by the stone during the last one second of motion is:
\[ \boxed{25 \, \text{m}} \]