Two bodies are projected simultaneously with the same velocity of `19.6 m//s` from the top of a tower one vertically upwards and the other vertically downwards. As they reach the ground, the time gap is

Options

A. 0 s
B. 2 s
C. 4 s
D. 6 s

Explanation

To solve the problem of two bodies projected from the top of a tower with the same initial velocity of \(19.6 \, \text{m/s}\), one going upwards and the other downwards, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Both bodies are projected with an initial velocity \(u = 19.6 \, \text{m/s}\).
- The acceleration due to gravity \(g = 9.8 \, \text{m/s}^2\) acts downwards.

2. **Equation of Motion for the Body Going Upwards**:
- For the body projected upwards, we can use the equation of motion:
\[ S = ut - \frac{1}{2}gt^2 \]
- At the highest point, the final velocity will be \(0\). Using the equation \(v = u - gt\), we can find the time taken to reach the highest point:
\[ 0 = 19.6 - 9.8t \implies t = 2 \, \text{s} \]
- The maximum height \(h\) reached by the body can be calculated using:
\[ h = ut - \frac{1}{2}gt^2 = 19.6(2) - \frac{1}{2}(9.8)(2^2) = 39.2 - 19.6 = 19.6 \, \text{m} \]

3. **Total Time for the Upward Body to Reach the Ground**:
- After reaching the maximum height, the body will fall back down. The total time taken to reach the ground will be the time to go up plus the time to come down.
- The time to fall from the maximum height \(h = 19.6 \, \text{m}\) can be calculated using:
\[ S = \frac{1}{2}gt^2 \implies 19.6 = \frac{1}{2}(9.8)t^2 \implies t^2 = 4 \implies t = 2 \, \text{s} \]
- Therefore, the total time for the upward body is:
\[ T_{\text{up}} = 2 + 2 = 4 \, \text{s} \]

4. **Equation of Motion for the Body Going Downwards**:
- For the body projected downwards, the equation of motion is:
\[ S = ut + \frac{1}{2}gt^2 \]
- The time taken for this body to reach the ground can be calculated as follows:
\[ S = 0 \, \text{(ground level)} \implies 0 = 19.6t + \frac{1}{2}(9.8)t^2 \]
- This is a quadratic equation in the form:
\[ 0 = 19.6t + 4.9t^2 \implies 4.9t^2 + 19.6t = 0 \implies t(4.9t + 19.6) = 0 \]
- The non-zero solution gives:
\[ t = -\frac{19.6}{4.9} = 4 \, \text{s} \]

5. **Calculate the Time Gap**:
- Both bodies take \(4 \, \text{s}\) to reach the ground. Therefore, the time gap between the two bodies is:
\[ \text{Time gap} = T_{\text{up}} - T_{\text{down}} = 4 - 4 = 0 \, \text{s} \]

### Final Answer:
The time gap between the two bodies when they reach the ground is \(0 \, \text{s}\).