Standing at a distance of 7 m from an 11.8 m high building, a...

The distance fallen by the kid before being caught is:

$h = 11.8 - 1.8 = 10 \text{ m}$

Using the second equation of motion for the freely falling kid ($u = 0$):

$h = ut + \dfrac{1}{2}gt^2$

$10 = 0 + \dfrac{1}{2} \times 9.8 \times t^2$

$10 = 4.9 t^2$

$t^2 = \dfrac{10}{4.9} = \dfrac{100}{49}$

$t = \dfrac{10}{7} \text{ s}$

The man must cover a horizontal distance of $7 \text{ m}$ in this time $t$ to catch the kid. Since his speed is uniform, the required speed $v$ is:

$v = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{7}{\dfrac{10}{7}}$

$v = \dfrac{49}{10} = 4.9 \text{ m/s}$