A block of mass 5 kg falls from the top of a building and hits the ground, 20 m below, 3 s later. Assuming the resistance experienced by the block is constant throughout the motion, find this resistance.

The mass of the block is $m = 5 \text{ kg}$.
The distance fallen is $s = 20 \text{ m}$.
The time taken is $t = 3 \text{ s}$.
The initial velocity is $u = 0 \text{ m/s}$ (since it falls from rest).
We will use the acceleration due to gravity $g = 10 \text{ m/s}^2$.
The kinematic equation relating distance, initial velocity, acceleration, and time is:

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

Since $u = 0$, the equation simplifies to $s = \frac{1}{2}at^2$.
Substitute the given values:

$$20 \text{ m} = (0 \text{ m/s})(3 \text{ s}) + \frac{1}{2}a(3 \text{ s})^2$$ $$20 \text{ m} = \frac{1}{2}a(9 \text{ s}^2)$$ $$40 \text{ m} = a(9 \text{ s}^2)$$ $$a = \frac{40}{9} \text{ m/s}^2$$ $$a \approx 4.44 \text{ m/s}^2$$

The forces acting on the block are the gravitational force ($F_g$) acting downwards and the resistance force ($R$) acting upwards. The net force ($F_{net}$) causes the acceleration $a$. Taking the downward direction as positive:

$$F_{net} = F_g - R$$

According to Newton's Second Law, $F_{net} = ma$.
So, $ma = F_g - R$.
The gravitational force is $F_g = mg$.

$$ma = mg - R$$

Rearrange the equation to solve for $R$:

$$R = mg - ma$$ $$R = m(g - a)$$

$$R = 5 \text{ kg} \left(10 \text{ m/s}^2 - \frac{40}{9} \text{ m/s}^2\right)$$ $$R = 5 \text{ kg} \left(\frac{90}{9} \text{ m/s}^2 - \frac{40}{9} \text{ m/s}^2\right)$$ $$R = 5 \text{ kg} \left(\frac{50}{9} \text{ m/s}^2\right)$$ $$R = \frac{250}{9} \text{ N}$$ $$R \approx 27.777... \text{ N}$$

Rounding to one decimal place, the resistance is:

$$R \approx 27.8 \text{ N}$$

The resistance experienced by the block is $27.8 \text{ N}$.