A bullet fired vertically up from the ground reaches a height 40 m in its path from the ground and it takes further time 2 seconds to reach the same point during descent. The total time of flight is `(g=10 ms^(-2))`

Options

A. 4 s
B. 3 s
C. 6 s
D. 8 s

Answer

To solve the problem, we need to find the total time of flight of a bullet fired vertically upwards that reaches a height of 40 m and takes 2 seconds to return to that height during its descent. We will use the equations of motion to find the required time.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The bullet is fired vertically upwards and reaches a height of 40 m.
- It takes 2 seconds to descend back to the same height of 40 m.

2. **Finding the Time to Reach Maximum Height**:
- Since it takes 2 seconds to come down to the height of 40 m, it must take 1 second to reach the maximum height (as the time taken to go up is equal to the time taken to come down).
- Therefore, the time taken to reach the maximum height is \( t_{up} = 1 \, \text{s} \).

3. **Using the First Equation of Motion**:
- We can use the first equation of motion:
\[ v = u + at \]
- At the maximum height, the final velocity \( v = 0 \) m/s, the initial velocity \( u \) is unknown, acceleration \( a = -g = -10 \, \text{m/s}^2 \), and time \( t = 1 \, \text{s} \).
- Plugging in the values:
\[ 0 = u - 10 \cdot 1 \]
- Rearranging gives:
\[ u = 10 \, \text{m/s} \]

4. **Finding the Initial Velocity at the Start**:
- Now we need to find the initial velocity \( u_0 \) when the bullet was fired. We can use the second equation of motion:
\[ v^2 = u_0^2 + 2as \]
- Here, \( v = 10 \, \text{m/s} \) (velocity at 40 m), \( a = -10 \, \text{m/s}^2 \), and \( s = 40 \, \text{m} \).
- Plugging in the values:
\[ (10)^2 = u_0^2 + 2(-10)(40) \]
- This simplifies to:
\[ 100 = u_0^2 - 800 \]
- Rearranging gives:
\[ u_0^2 = 900 \quad \Rightarrow \quad u_0 = \sqrt{900} = 30 \, \text{m/s} \]

5. **Calculating Total Time of Flight**:
- The total time of flight \( T \) is given by the time taken to reach the maximum height and the time taken to descend back to the ground.
- The time to reach maximum height is \( t_{up} = \frac{u_0}{g} = \frac{30}{10} = 3 \, \text{s} \).
- Therefore, the total time of flight is:
\[ T = 2 \times t_{up} = 2 \times 3 = 6 \, \text{s} \]

### Final Answer:
The total time of flight is \( 6 \, \text{s} \).

---