A person in a lift which descends down with an acceleration of `1.8 m//s^2` drops a stone from a height of 2 m. The time of decent is

Options

A. `sqrt2` sec
B. `2sqrt2` sec
C. `1//sqrt2` sec
D. `1/2` sec

Answer

To solve the problem step by step, we will analyze the motion of the stone dropped from a height of 2 meters while the lift is descending with an acceleration of 1.8 m/s².

### Step 1: Understand the Forces Acting on the Stone
When the stone is dropped, it experiences two forces:
1. The gravitational force (weight) acting downwards, which is \( mg \).
2. A pseudo force acting upwards due to the downward acceleration of the lift, which is \( ma \), where \( a \) is the acceleration of the lift.

### Step 2: Calculate the Net Acceleration of the Stone
The net acceleration of the stone with respect to the lift can be calculated as follows:
\[ a_{\text{net}} = g - a \]
Where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( a = 1.8 \, \text{m/s}^2 \) (acceleration of the lift)

Substituting the values:
\[ a_{\text{net}} = 9.8 \, \text{m/s}^2 - 1.8 \, \text{m/s}^2 = 8.0 \, \text{m/s}^2 \]

### Step 3: Use the Second Equation of Motion
We will use the second equation of motion to find the time of descent. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = 2 \, \text{m} \) (displacement)
- \( u = 0 \, \text{m/s} \) (initial velocity, since the stone is dropped)
- \( a = 8.0 \, \text{m/s}^2 \) (net acceleration)

Substituting the known values into the equation:
\[ 2 = 0 \cdot t + \frac{1}{2} \cdot 8.0 \cdot t^2 \]
This simplifies to:
\[ 2 = 4t^2 \]

### Step 4: Solve for Time \( t \)
Rearranging the equation gives:
\[ t^2 = \frac{2}{4} = \frac{1}{2} \]
Taking the square root of both sides:
\[ t = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \, \text{s} \]

### Final Answer
The time of descent for the stone is:
\[ t = \frac{1}{\sqrt{2}} \, \text{s} \]

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