A healthy youngman standing at a distance of 7 m from 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms hieght (1.8m)?

Explanation

To solve the problem, we need to determine the speed at which the young man should run to catch the kid falling from the building. Here’s a step-by-step solution:

### Step 1: Determine the height from which the kid falls
The total height of the building is 11.8 m, and the height at which the young man can catch the kid is 1.8 m. Therefore, the distance the kid falls is:

\[ \text{Distance fallen} = \text{Height of the building} - \text{Height at which the young man catches the kid} \]
\[ \text{Distance fallen} = 11.8 \, \text{m} - 1.8 \, \text{m} = 10 \, \text{m} \]

### Step 2: Calculate the time taken for the kid to fall
We can use the second equation of motion to find the time taken for the kid to fall 10 m under gravity. The equation is:

\[ S = ut + \frac{1}{2}gt^2 \]

Where:
- \( S \) = distance fallen = 10 m
- \( u \) = initial velocity = 0 (the kid starts from rest)
- \( g \) = acceleration due to gravity ≈ 9.81 m/s²
- \( t \) = time in seconds

Substituting the values into the equation:

\[ 10 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \]
\[ 10 = 4.905t^2 \]
\[ t^2 = \frac{10}{4.905} \approx 2.04 \]
\[ t \approx \sqrt{2.04} \approx 1.43 \, \text{s} \]

### Step 3: Calculate the speed required for the young man to run
The young man is standing 7 m away from the building. To catch the kid, he must cover this distance in the same time it takes for the kid to fall (1.43 s). The speed \( v \) can be calculated using the formula:

\[ v = \frac{\text{Distance}}{\text{Time}} \]
\[ v = \frac{7 \, \text{m}}{1.43 \, \text{s}} \approx 4.90 \, \text{m/s} \]

### Final Answer
The speed at which the young man should run to catch the kid is approximately **4.90 m/s**.

---