A body X is projected upwards with a velocity of `98 ms^(-1)`, after 4s, a second body Y is also projected upwards with the same initial velocity . Two bodies will meet after

Options

A. 8 s
B. 10 s
C. 12 s
D. 14 s

Explanation

To solve the problem, we will analyze the motion of both bodies X and Y, which are projected upwards with the same initial velocity. We will use the equations of motion to find out when they meet.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Initial velocity (u) = 98 m/s (for both bodies)
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)
- Time of projection for body X (t) = t seconds
- Time of projection for body Y = t - 4 seconds (since it is projected 4 seconds later)

2. **Equation of Motion for Body X:**
The displacement of body X after time t can be given by the equation:
\[ s_X = u t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_X = 98t - \frac{1}{2} \cdot 9.8 \cdot t^2 = 98t - 4.9t^2 \]

3. **Equation of Motion for Body Y:**
The displacement of body Y after (t - 4) seconds can be given by:
\[ s_Y = u(t - 4) - \frac{1}{2} g (t - 4)^2 \]
Substituting the values:
\[ s_Y = 98(t - 4) - \frac{1}{2} \cdot 9.8 (t - 4)^2 \]
Simplifying:
\[ s_Y = 98t - 392 - 4.9(t^2 - 8t + 16) = 98t - 392 - 4.9t^2 + 39.2t - 78.4 \]
\[ s_Y = 98t - 4.9t^2 + 39.2t - 470.4 = 137.2t - 4.9t^2 - 470.4 \]

4. **Setting the Displacements Equal:**
Since both bodies meet at the same displacement, we set \(s_X = s_Y\):
\[ 98t - 4.9t^2 = 137.2t - 4.9t^2 - 470.4 \]

5. **Simplifying the Equation:**
Cancel out \(4.9t^2\) from both sides:
\[ 98t = 137.2t - 470.4 \]
Rearranging gives:
\[ 470.4 = 137.2t - 98t \]
\[ 470.4 = 39.2t \]

6. **Solving for t:**
\[ t = \frac{470.4}{39.2} = 12 \text{ seconds} \]

### Conclusion:
The two bodies will meet after **12 seconds**.