Height Where Magnitude of Velocity Equals Acceleration

To find the height above the ground at which the magnitude of the velocity equals the magnitude of acceleration due to gravity, we proceed step-by-step using standard kinematic equations.

1. Given Data:

• Initial height above the ground, $H = 18\text{ m}$
• Initial velocity of the ball, $u = 0\text{ m/s}$ (dropped from rest)
• Acceleration due to gravity, $g = 10\text{ m/s}^2$

2. Condition for Velocity: We are given that the magnitude of velocity $|v|$ is equal to the magnitude of acceleration due to gravity $g$ (in the same set of units): $|v| = g = 10\text{ m/s}$

3. Calculating Distance Fallen: Let $y$ be the distance fallen from the release point. Using the third equation of motion: $v^2 = u^2 + 2gy$

Substituting $u = 0$ and $|v| = 10\text{ m/s}$: $(10)^2 = 0 + 2(10)y$ $100 = 20y$ $y = 5\text{ m}$

4. Finding Height Above the Ground: The height above the ground, $h$, is given by subtracting the distance fallen from the initial height: $h = H - y$ $h = 18\text{ m} - 5\text{ m} = 13\text{ m}$

Thus, the height above the ground is 13.