Given:

• Mass of ball $m = 0.5\,\text{kg}$ (not needed here)
• Initial height $= 10\,\text{m}$
• Ball is dropped, so initial velocity $u = 0$
• Acceleration due to gravity $g = 10\,\text{m/s}^2$

Condition to be satisfied: The magnitude of velocity becomes equal to the magnitude of acceleration due to gravity.

So, $v = g = 10\,\text{m/s}$

Use equation of motion: $v^2 = u^2 + 2gs$ Since $u=0$, $v^2 = 2gs$

Substitute $v=10$ and $g=10$: $10^2 = 2(10)s$ $100 = 20s$ $s = 5\,\text{m}$

Here, $s$ is the distance fallen from the top.

Find the height above the ground: Initial height is $10\,\text{m}$, so remaining height is $h = 10 - 5 = 5\,\text{m}$

Final Answer: The required height is $\boxed{5\,\text{m}}$