A healthy young man stand at a distance of $7m$ from a $11.8m$height building sees a kid slipping from the top floor. With what speed (assume uniform) should he run to catch the kid at the arm s height at $1.8m$

Step 1. Given data

A healthy young man stand at a distance of $7m$.

The height of the building is $11.8m$.

The kid should caught over $1.8m$ from the ground level.

Initial velocity is $0m/s$

Step 2. Formula used

Speed =$\frac{s}{t}$

where $s$ is the distance and $t$ is the time

Equation of the motion

$s=ut+\frac{1}{2}g{t}^2$

where $s$is the distance , $u$ the initial velocity $g$is the gravitational constant.(take $g=9.8m/s^2$) and $t$ is the time.

Step 3. Find the distance traveled by the kid

The distance covered by the kid before he was caught is $=11.8-1.8=10m$

Step 4. Find the time taken to fall

Let the time taken to fall is $t$

$$\begin{gathered} s=10m \\ g=9.8m/s^2 \\ u=0m/s \end{gathered}$$

The equation of the motion is

$$\begin{gathered} s=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow 10=0+\frac{1}{2}\times 9.8\times t^2 \\ \Rightarrow t^2=2.04 \\ \Rightarrow t=1.42s \end{gathered}$$

Step 5. Find the speed at which the man ran

The speed at which the man ran in order to catch the kid is

Speed$=\frac{s}{t}$

$$\begin{gathered} =\frac{7}{1.42} \\ =4.9m/s \end{gathered}$$

Hence, the man ran at a speed of $4.9m/s$.