A ball is dropped onto a sandy floor from a height of 5 m and...

The velocity of the ball just before entering the sand is given by the equation of motion:

$v^2 = u^2 + 2gh$

Since the ball is dropped, its initial velocity $u = 0$. Taking $g = 9.8 \text{ m/s}^2$ and $h = 5 \text{ m}$:

$v^2 = 0 + 2 \times 9.8 \times 5 = 98 \text{ m}^2/\text{s}^2$

Let $a$ be the uniform retardation of the ball inside the sand. The final velocity $v_f$ becomes zero after penetrating a distance $s = 10 \text{ cm} = 0.1 \text{ m}$.

Using the third equation of motion for the motion inside the sand:

$v_f^2 = v^2 - 2as$

Substituting the values:

$0 = 98 - 2 \times a \times 0.1$

$0.2a = 98$

$a = \dfrac{98}{0.2} = 490 \text{ m/s}^2$