A stone is dropped from rest from the top of a tower 19.6m high. The distance travelled during the last second of its fall is (given `g=9.8m//s^(2)`):

Options

A. 9.8m
B. 14.7m
C. 4.9m
D. 19.6m

Explanation

To solve the problem of finding the distance traveled by a stone during the last second of its fall from a tower of height 19.6 m, we can follow these steps:

### Step 1: Determine the time of fall
We start with the equation of motion for an object in free fall:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the distance fallen (19.6 m),
- \( u \) is the initial velocity (0 m/s, since the stone is dropped),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( t \) is the time in seconds.

Substituting the known values into the equation:
\[ 19.6 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ 19.6 = 4.9t^2 \]
Now, we can solve for \( t^2 \):
\[ t^2 = \frac{19.6}{4.9} = 4 \]
Taking the square root gives:
\[ t = 2 \text{ seconds} \]

### Step 2: Calculate the distance traveled in the last second
Now that we know the total time of fall is 2 seconds, we need to find the distance traveled during the last second, which is from \( t = 1 \) second to \( t = 2 \) seconds.

We can find the distance fallen in the first 2 seconds:
\[ s_{2} = ut + \frac{1}{2}gt^2 \]
Substituting \( t = 2 \) seconds:
\[ s_{2} = 0 \cdot 2 + \frac{1}{2} \cdot 9.8 \cdot (2)^2 \]
\[ s_{2} = 0 + \frac{1}{2} \cdot 9.8 \cdot 4 = 19.6 \text{ m} \]

Next, we calculate the distance fallen in the first second:
\[ s_{1} = ut + \frac{1}{2}gt^2 \]
Substituting \( t = 1 \) second:
\[ s_{1} = 0 \cdot 1 + \frac{1}{2} \cdot 9.8 \cdot (1)^2 \]
\[ s_{1} = 0 + \frac{1}{2} \cdot 9.8 \cdot 1 = 4.9 \text{ m} \]

### Step 3: Calculate the distance traveled during the last second
The distance traveled during the last second is the difference between the total distance fallen in 2 seconds and the distance fallen in the first second:
\[ \text{Distance in last second} = s_{2} - s_{1} \]
\[ \text{Distance in last second} = 19.6 - 4.9 = 14.7 \text{ m} \]

Thus, the distance traveled during the last second of its fall is **14.7 m**.