A stone is dropped from the top of a tower of height 125 m. The distance travelled by it during last second of its fall is `(g=10 ms^(-2))`

Options

A. 22.5 m
B. 45 m
C. 75 m
D. 90 m

Answer

To solve the problem of finding the distance traveled by a stone during the last second of its fall from a height of 125 m, we will follow these steps:

### Step 1: Determine the time of flight
We will use the second equation of motion to find the time of flight. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = 125 \, \text{m} \) (height of the tower)
- \( u = 0 \, \text{m/s} \) (initial velocity, since the stone is dropped)
- \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values into the equation:
\[ 125 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 125 = 5t^2 \]

Now, solving for \( t^2 \):
\[ t^2 = \frac{125}{5} = 25 \]
Taking the square root:
\[ t = \sqrt{25} = 5 \, \text{s} \]

### Step 2: Calculate the distance traveled during the last second
To find the distance traveled during the last second (the 5th second), we can use the formula:
\[ s_n = u + \frac{a}{2} \cdot (2n - 1) \]
Where:
- \( n = 5 \) (since we are looking for the distance in the 5th second)
- \( u = 0 \, \text{m/s} \)
- \( a = 10 \, \text{m/s}^2 \)

Substituting the values into the equation:
\[ s_5 = 0 + \frac{10}{2} \cdot (2 \cdot 5 - 1) \]
This simplifies to:
\[ s_5 = 5 \cdot (10 - 1) = 5 \cdot 9 = 45 \, \text{m} \]

### Final Answer
The distance traveled by the stone during the last second of its fall is **45 meters**.

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