Question

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 air resistance)}

Answer 1

Correct Answer: 13

Step 1: Understanding the Question:
A ball is dropped from a height \(H_{total} = 18 \text{ m}\). We need to find the height \(H\) above the ground where the velocity \(v\) reaches a magnitude equal to the acceleration due to gravity \(g = 10 \text{ m/s}^2\).
Step 2: Key Formula or Approach:
1. Third equation of motion: \(v^2 = u^2 + 2gh\), where \(u\) is initial velocity and \(h\) is the distance fallen.
2. Since it is dropped from rest, \(u = 0\).
3. The distance fallen \(h\) is related to height above ground \(H\) by \(h = H_{total} - H\).
Step 3: Detailed Explanation:
The condition given is \(|v| = |g| = 10 \text{ m/s}\).
Substitute this into the equation of motion:
\[ v^2 = 0 + 2gh \]
\[ 10^2 = 2 \times 10 \times h \]
\[ 100 = 20h \]
\[ h = \frac{100}{20} = 5 \text{ m} \]
This \(h\) is the distance the ball has fallen from the starting point.
The height above the ground (\(H\)) is:
\[ H = H_{total} - h \]
\[ H = 18 \text{ m} - 5 \text{ m} = 13 \text{ m} \]
Step 4: Final Answer:
The height above the ground is \(13 \text{ m}\).