A lift whose cage is 3 m high is moving up with an acceleration of `2m//s^(2)`. A piece of stone is dropped from the top of the cage of the lift when its velocity is 8m/s. if `g=10m//s^(2)`, then the stone will reach the floor of the lift after

Options

A. 0.7s
B. 0.5s
C. 0.4s
D. 0.3s

Explanation

To solve the problem step by step, we will analyze the motion of the stone relative to the lift and use the equations of motion.

### Step 1: Understand the scenario
The lift is moving upwards with an acceleration of \(2 \, \text{m/s}^2\), and the stone is dropped from the top of the lift when the lift's velocity is \(8 \, \text{m/s}\). The height of the lift is \(3 \, \text{m}\), and we need to find out how long it takes for the stone to reach the floor of the lift.

### Step 2: Establish the relative motion
When the stone is dropped, it has the same upward velocity as the lift, which is \(8 \, \text{m/s}\). In the frame of reference of the lift, the initial velocity of the stone is \(0 \, \text{m/s}\) because both the lift and the stone are moving upwards at the same speed.

### Step 3: Determine the effective acceleration
In the lift's frame, the stone experiences a downward acceleration due to gravity and the upward acceleration of the lift. The effective acceleration \(a\) acting on the stone is:
\[ a = g + a_{\text{lift}} = 10 \, \text{m/s}^2 + 2 \, \text{m/s}^2 = 12 \, \text{m/s}^2 \]

### Step 4: Use the equation of motion
We can use the second equation of motion to find the time \(t\) it takes for the stone to fall \(3 \, \text{m}\):
\[ S = ut + \frac{1}{2} a t^2 \]
Here, \(S = 3 \, \text{m}\), \(u = 0 \, \text{m/s}\), and \(a = 12 \, \text{m/s}^2\).

Substituting the values:
\[ 3 = 0 \cdot t + \frac{1}{2} \cdot 12 \cdot t^2 \]
\[ 3 = 6t^2 \]
\[ t^2 = \frac{3}{6} = \frac{1}{2} \]
\[ t = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \, \text{s} \]

### Step 5: Conclusion
The stone will reach the floor of the lift after approximately \(0.707 \, \text{s}\).

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