Starting from rest, A lift moves up with an acceleration of `2 ms^(-2)`. In this lift , a ball is dropped from a height of `1.5m` (with respect to floor of lift). The time taken by the ball to reach the floor of the lift is (`g=10 ms^(-2)`)

Options

A. `0.3 s`
B. `0.2 s`
C. `0.16 s `
D. `0.5 s`

Explanation

To solve the problem step by step, we need to analyze the motion of the ball dropped from a height of 1.5 m inside an accelerating lift.

### Step 1: Understand the forces acting on the ball
When the ball is dropped, it experiences two main forces:
1. The gravitational force acting downwards, which is \( mg \).
2. The pseudo force acting upwards due to the acceleration of the lift, which is \( ma \).

### Step 2: Determine the effective acceleration of the ball
The effective acceleration of the ball can be calculated by combining the gravitational acceleration \( g \) and the lift's acceleration \( a \):
\[ \text{Effective acceleration} = g + a \]
Given \( g = 10 \, \text{m/s}^2 \) and \( a = 2 \, \text{m/s}^2 \):
\[ \text{Effective acceleration} = 10 + 2 = 12 \, \text{m/s}^2 \]

### Step 3: Use the kinematic equation to find the time taken
We will use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = 1.5 \, \text{m} \) (the height from which the ball is dropped),
- \( u = 0 \, \text{m/s} \) (initial velocity of the ball),
- \( a = 12 \, \text{m/s}^2 \) (effective acceleration),
- \( t \) is the time taken to reach the floor of the lift.

Substituting the known values into the equation:
\[ 1.5 = 0 \cdot t + \frac{1}{2} \cdot 12 \cdot t^2 \]
This simplifies to:
\[ 1.5 = 6t^2 \]

### Step 4: Solve for \( t^2 \)
Rearranging the equation gives:
\[ t^2 = \frac{1.5}{6} = 0.25 \]

### Step 5: Calculate \( t \)
Taking the square root of both sides:
\[ t = \sqrt{0.25} = 0.5 \, \text{s} \]

### Final Answer
The time taken by the ball to reach the floor of the lift is \( 0.5 \, \text{s} \).

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