A mango fruit drops from a height of 10 m. What is its velocity just before hitting the ground? (Assume g = 10 m/s²)

The mango fruit drops, so its initial velocity is $u = 0 \, \text{m/s}$.
The height from which it drops is the displacement, $s = 10 \, \text{m}$.
The acceleration due to gravity is $a = g = 10 \, \text{m/s}^2$ (a common approximation in such problems, which aligns with the given options).
We need to find the final velocity, $v$, just before hitting the ground.
The relevant kinematic equation is:

$$v^2 = u^2 + 2as$$

$$v^2 = (0 \, \text{m/s})^2 + 2(10 \, \text{m/s}^2)(10 \, \text{m})$$ $$v^2 = 0 + 200 \, \text{m}^2/\text{s}^2$$ $$v^2 = 200 \, \text{m}^2/\text{s}^2$$

Take the square root of both sides:

$$v = \sqrt{200} \, \text{m/s}$$

Simplify the square root:

$$v = \sqrt{100 \times 2} \, \text{m/s}$$ $$v = 10\sqrt{2} \, \text{m/s}$$

The calculated velocity is $10\sqrt{2} \, \text{m/s}$, which matches option A.

The final answer is $\text{A. } 10\sqrt{2} \, \text{ms}^{-1}$.