From \(18\) m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _____ m.
(Take \(g = 10\) m/s\(^2\) and neglect the air resistance)

Let the ball fall a distance \(s\) from the initial height.

Initial velocity \(u = 0\)

Acceleration \(a = g = 10\) m/s\(^2\)

Given that the magnitude of velocity is equal to the magnitude of acceleration due to gravity, we have \(v = 10\) m/s.

Using the equation of motion \(v^2 = u^2 + 2gs\):

\((10)^2 = 0^2 + 2 \times 10 \times s\)

\(100 = 20s\)

\(s = 5\) m

The distance fallen by the ball is \(5\) m.

The height above the ground is \(h = H - s\)

\(h = 18 - 5 = 13\) m

Answer: \(13\)