A stuntman plans to run across a roof top and then horizontally off it to land on the roof of next bulding. The roof of the next building is 4.9 metre below the first one and 6.2 metre away from it. What should be his minimum roof top speed in m/s. so that he can successfully make the jump?

Options

A. (a)`3.1`
B. (b)`4.0`
C. (c)`4.9`
D. (d)`6.2`

Answer

To solve the problem of the stuntman jumping from one rooftop to another, we need to determine the minimum speed he must have when he leaves the first rooftop to successfully land on the second rooftop.

### Step-by-Step Solution:

1. **Identify the vertical drop and horizontal distance**:
- The vertical distance (drop) to the next building is given as \( s = 4.9 \) meters.
- The horizontal distance to the next building is \( d = 6.2 \) meters.

2. **Use the equations of motion to find the time of fall**:
- The equation of motion for vertical displacement is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, the initial vertical velocity \( u = 0 \) (since the stuntman jumps horizontally), and the acceleration \( a = g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
- Substituting these values into the equation, we get:
\[ s = 0 \cdot t + \frac{1}{2} g t^2 \]
\[ s = \frac{1}{2} g t^2 \]
- Rearranging for \( t \):
\[ t^2 = \frac{2s}{g} \]
\[ t = \sqrt{\frac{2s}{g}} \]

3. **Calculate the time taken to fall 4.9 meters**:
- Substitute \( s = 4.9 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ t = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1} = 1 \, \text{s} \]

4. **Calculate the minimum horizontal speed**:
- The horizontal distance \( d = 6.2 \, \text{m} \) must be covered in the time \( t = 1 \, \text{s} \).
- The formula for speed is:
\[ v = \frac{d}{t} \]
- Substituting the values:
\[ v = \frac{6.2 \, \text{m}}{1 \, \text{s}} = 6.2 \, \text{m/s} \]

5. **Conclusion**:
- The minimum rooftop speed required for the stuntman to successfully make the jump is \( 6.2 \, \text{m/s} \).

### Final Answer:
The minimum rooftop speed required is **6.2 m/s**.