A ball is dropped from rest from a height of 18 m above the ground. Find the height above the ground at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity.

The ball is released from rest, so:

u = 0

The acceleration acting on the ball is:

g = 10 m/s²

According to the problem, the magnitude of velocity becomes equal to g.

Therefore:

v = 10 m/s

Using the equation of motion:

v² = u² + 2gs

Substituting values:

10² = 0 + 2(10)s

100 = 20s

s = 5 m

This means the ball has fallen 5 m from its initial position.

The initial height above the ground was:

18 m

Hence, the remaining height above the ground is:

18 − 5 = 13 m

This problem illustrates the relationship between displacement and velocity in uniformly accelerated motion. Since gravity provides constant acceleration, the velocity increases steadily as the object falls.

Therefore, the required height above the ground is:

13 m