A freely falling body travels _____ of total distance in 5th second

Options

A. `8%`
B. `12%`
C. `25%`
D. `36%`

Answer

To solve the problem of finding the distance traveled by a freely falling body in the fifth second as a percentage of the total distance traveled in five seconds, we can follow these steps:

### Step 1: Understand the problem
We need to find the distance traveled in the fifth second and express it as a percentage of the total distance traveled in five seconds.

### Step 2: Use the second equation of motion
The second equation of motion is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance traveled,
- \( u \) is the initial velocity (which is 0 for a freely falling body),
- \( a \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.

### Step 3: Calculate the distance traveled in the first four seconds
For \( t = 4 \) seconds:
\[ s_4 = 0 \cdot 4 + \frac{1}{2} \cdot 10 \cdot (4^2) \]
\[ s_4 = 0 + \frac{1}{2} \cdot 10 \cdot 16 \]
\[ s_4 = 80 \, \text{meters} \]

### Step 4: Calculate the distance traveled in five seconds
For \( t = 5 \) seconds:
\[ s_5 = 0 \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5^2) \]
\[ s_5 = 0 + \frac{1}{2} \cdot 10 \cdot 25 \]
\[ s_5 = 125 \, \text{meters} \]

### Step 5: Calculate the distance traveled in the fifth second
The distance traveled in the fifth second (\( s_5 - s_4 \)):
\[ \text{Distance in 5th second} = s_5 - s_4 = 125 - 80 = 45 \, \text{meters} \]

### Step 6: Calculate the percentage of the distance traveled in the fifth second
To find the percentage of the distance traveled in the fifth second relative to the total distance:
\[ \text{Percentage} = \left( \frac{\text{Distance in 5th second}}{\text{Total distance}} \right) \times 100 \]
\[ \text{Percentage} = \left( \frac{45}{125} \right) \times 100 \]
\[ \text{Percentage} = 36\% \]

### Final Answer
The freely falling body travels **36%** of the total distance in the fifth second.

---