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:
The topic of this question is Kinematics (Motion in a Straight Line).
We are asked to find the specific height of a falling object at the moment its speed becomes numerically equal to the acceleration due to gravity.
Step 2: Key Formula or Approach:
We will use the third equation of motion for an object falling under gravity from rest:
\[ v^2 = u^2 + 2gh \]
Since the ball is dropped from rest, the initial velocity \(u = 0\). The equation simplifies to \(v^2 = 2gh\), where \(h\) is the distance the ball has fallen.
Step 3: Detailed Explanation:
The condition given in the question is that the magnitude of velocity \(|v|\) equals the magnitude of acceleration \(|g|\).
So, we set \(v = g = 10 \text{ m/s}\).
Substitute this value into the simplified equation of motion to find the distance fallen, \(h\):
\[ (10)^2 = 2 \times (10) \times h \]
\[ 100 = 20h \]
\[ h = \frac{100}{20} = 5 \text{ m} \]
This value, \(h = 5 \text{ m}\), represents the distance the ball has fallen from the top.
The question asks for the height above the ground, which is the initial height minus the distance fallen:
\[ \text{Height above ground} = \text{Total height} - h = 18 \text{ m} - 5 \text{ m} = 13 \text{ m} \]
Step 4: Final Answer:
The height above the ground is \(13 \text{ m}\).