A bullet fired vertically up from the ground reaches a point in its path after 3s and 13 s, both times being taken from the instant of firing the bullet. The velocity of projection of the bullet is `(g=10 ms^(-2))`

Options

A. `80ms^(-1)`
B. `40 ms^(-1)`
C. `160 ms^(-1)`
D. `20 ms^(-1)`

Answer

To find the velocity of projection of the bullet, we can use the information given in the problem. The bullet reaches a certain point in its path at two different times: 3 seconds and 13 seconds after being fired. We also know that the acceleration due to gravity (g) is 10 m/s².

### Step-by-Step Solution:

1. **Identify the times**:
- Let \( t_1 = 3 \) seconds
- Let \( t_2 = 13 \) seconds

2. **Use the formula for velocity of projection**:
The formula relating the initial velocity (u), gravity (g), and the times at which the bullet reaches the same height is:
\[ u = \frac{g}{2} (t_1 + t_2) \]

3. **Substitute the values into the formula**:
- Substitute \( g = 10 \, \text{m/s}^2 \), \( t_1 = 3 \, \text{s} \), and \( t_2 = 13 \, \text{s} \):
\[ u = \frac{10}{2} (3 + 13) \]

4. **Calculate the sum of times**:
- Calculate \( t_1 + t_2 \):
\[ t_1 + t_2 = 3 + 13 = 16 \, \text{s} \]

5. **Calculate the initial velocity**:
- Now substitute back into the equation:
\[ u = \frac{10}{2} \times 16 \]
\[ u = 5 \times 16 = 80 \, \text{m/s} \]

6. **Final result**:
- The velocity of projection of the bullet is \( 80 \, \text{m/s} \).

### Final Answer:
The velocity of projection of the bullet is \( 80 \, \text{m/s} \).