A body is droped from a height 122.5 m. If its stopped after 3 seconds and again released the further time of descent is `(g=9.8 "m/s"^2)`

Options

A. 2 s
B. 3 s
C. 4 s
D. 5 s

Answer

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the problem
A body is dropped from a height of 122.5 m. It falls for 3 seconds before being stopped and then released again. We need to find out how much longer it takes to reach the ground after being released again.

### Step 2: Calculate the distance fallen in the first 3 seconds
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) = distance fallen
- \( u \) = initial velocity (which is 0 m/s since the body is dropped)
- \( a \) = acceleration due to gravity (9.8 m/s²)
- \( t \) = time (3 seconds)

Substituting the values:
\[ s_1 = 0 \cdot 3 + \frac{1}{2} \cdot 9.8 \cdot (3^2) \]
\[ s_1 = 0 + \frac{1}{2} \cdot 9.8 \cdot 9 \]
\[ s_1 = 4.9 \cdot 9 \]
\[ s_1 = 44.1 \text{ m} \]

### Step 3: Calculate the remaining height
The initial height is 122.5 m, and the distance fallen in the first 3 seconds is 44.1 m. Therefore, the remaining height \( s_2 \) is:
\[ s_2 = 122.5 - 44.1 \]
\[ s_2 = 78.4 \text{ m} \]

### Step 4: Calculate the time taken to fall the remaining height
Now, we need to find the time taken to fall the remaining height of 78.4 m after being released again. We will use the same equation of motion:
\[ s_2 = ut + \frac{1}{2} a t^2 \]

Where:
- \( s_2 = 78.4 \text{ m} \)
- \( u = 0 \text{ m/s} \) (the body is released from rest)
- \( a = 9.8 \text{ m/s}^2 \)
- \( t \) is what we need to find.

Substituting the values:
\[ 78.4 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ 78.4 = \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ 78.4 = 4.9 \cdot t^2 \]

### Step 5: Solve for \( t^2 \)
Rearranging gives:
\[ t^2 = \frac{78.4}{4.9} \]
\[ t^2 = 16 \]

### Step 6: Calculate \( t \)
Taking the square root:
\[ t = \sqrt{16} \]
\[ t = 4 \text{ seconds} \]

### Final Answer
The total time of descent after being released again is **4 seconds**.

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