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 due to gravity is ______ m. (Take g

Let's start with the theory for Free Fall.

When an object is in free fall, it moves under the influence of gravity alone, neglecting air resistance. The acceleration is constant and equal to the acceleration due to gravity, $g$. We typically take $g = 9.8 \text{ m/s}^2$ or $10 \text{ m/s}^2$ as specified.

The equations of motion for constant acceleration are:

1. $v = u + at$
2. $s = ut + \frac{1}{2}at^2$
3. $v^2 = u^2 + 2as$

For free fall, if we consider downward as positive:

• Initial velocity $u$. If dropped from rest, $u=0$.
• Acceleration $a = g$.
• Displacement $s$ becomes height $h$.
• Final velocity $v$.

The equations become:

1. $v = u + gt$
2. $h = ut + \frac{1}{2}gt^2$
3. $v^2 = u^2 + 2gh$

Now, try solving question 1:

1. From $18 \text{ m}$ 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 ______ $\text{m}$. (Take $g = 10 \text{ m/s}^2$ and neglect the air resistance)