A body projected vertically upwards with a velocity u returns to the starting point in 4 seconds. If `g = 10m//sec`, the value of u is

Options

A. `5m//sec`
B. `10m//sec`
C. `15m//sec`
D. `20m//sec`

Answer

To solve the problem of a body projected vertically upwards with an initial velocity \( u \) that returns to the starting point in 4 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
The total time for the body to go up and come back down is 4 seconds. Since the time taken to reach the maximum height is equal to the time taken to return to the starting point, the time to reach the maximum height is:
\[ t = \frac{4 \text{ seconds}}{2} = 2 \text{ seconds} \]

2. **Using the Equation of Motion**:
At the maximum height, the final velocity \( v \) is 0 m/s. We can use the first equation of motion:
\[ v = u + at \]
Here, \( v = 0 \) (at maximum height), \( a = -g \) (acceleration due to gravity acting downwards), and \( t = 2 \text{ seconds} \).

3. **Substituting Values**:
Substituting the known values into the equation:
\[ 0 = u - g \cdot t \]
Rearranging gives:
\[ u = g \cdot t \]

4. **Calculating the Initial Velocity \( u \)**:
Given \( g = 10 \text{ m/s}^2 \) and \( t = 2 \text{ seconds} \):
\[ u = 10 \text{ m/s}^2 \cdot 2 \text{ seconds} = 20 \text{ m/s} \]

5. **Final Answer**:
Therefore, the value of \( u \) is:
\[ u = 20 \text{ m/s} \]