A body is projected vertically upwards wth a velocity `u=5m//s`. After time `t` another body is projected vertically upward from the same point with a velocity `v=3m//s`. If they meet in minimum time duration measured from the projection of first body, then `t=k/g` sec find `k` (where `g` is gravitatio acceleration).

Explanation

To solve the problem, we need to find the time \( t \) at which two bodies projected vertically upwards meet, and express it in the form \( t = \frac{k}{g} \), where \( g \) is the gravitational acceleration.

### Step-by-Step Solution

1. **Identify the Variables**:
- The first body is projected with an initial velocity \( u = 5 \, \text{m/s} \).
- The second body is projected after time \( t \) with an initial velocity \( v = 3 \, \text{m/s} \).
- Let \( g \) be the acceleration due to gravity.

2. **Determine the Time of Flight**:
- Let \( T \) be the total time from the projection of the first body until they meet.
- The time for the first body to reach the height \( h \) is \( T \).
- The time for the second body to reach the same height \( h \) is \( T - t \).

3. **Use the Equation of Motion**:
- For the first body:
\[ h = uT - \frac{1}{2}gT^2 \]
Substituting \( u = 5 \):
\[ h = 5T - \frac{1}{2}gT^2 \]
- For the second body:
\[ h = v(T - t) - \frac{1}{2}g(T - t)^2 \]
Substituting \( v = 3 \):
\[ h = 3(T - t) - \frac{1}{2}g(T - t)^2 \]

4. **Set the Heights Equal**:
\[ 5T - \frac{1}{2}gT^2 = 3(T - t) - \frac{1}{2}g(T - t)^2 \]

5. **Expand and Simplify**:
- Expanding the right side:
\[ 3T - 3t - \frac{1}{2}g(T^2 - 2Tt + t^2) \]
- Equating both sides:
\[ 5T - \frac{1}{2}gT^2 = 3T - 3t - \frac{1}{2}gT^2 + gTt - \frac{1}{2}gt^2 \]
- Rearranging gives:
\[ 2T + 3t = gTt - \frac{1}{2}gt^2 \]

6. **Rearranging the Equation**:
- Collecting terms leads to:
\[ 5T - 3T + 3t = gTt - \frac{1}{2}gt^2 \]
\[ 2T + 3t = gTt - \frac{1}{2}gt^2 \]

7. **Finding Minimum Time**:
- To minimize \( T \), we differentiate the equation with respect to \( t \) and set it to zero.
- After differentiating and simplifying, we find:
\[ 50t^2 - 20t - 6 = 0 \]
- Solving this quadratic equation gives:
\[ t = 0.6 \, \text{s} \quad (\text{rejecting negative root}) \]

8. **Expressing in Terms of \( g \)**:
- Given \( t = \frac{k}{g} \) and substituting \( g = 10 \, \text{m/s}^2 \):
\[ 0.6 = \frac{k}{10} \]
\[ k = 6 \]

### Final Answer
The value of \( k \) is \( \boxed{6} \).