A very broad elevator is going up vertically with a constant acceleration of `2m//s^(2)`. At the instant when its velocity is `4m//s`, a ball is projected from the floor of the lift wht as of `4m//s` relative to the floor at an elevation of `30^(@)`. The time taken by the ball to return the floor is `(g=10m//s^(2))`

Options

A. `1/2s`
B. `1/3s`
C. `1/4s`
D. `1s`

Answer

To solve the problem, we need to analyze the motion of the ball projected from the elevator. Here’s a step-by-step solution:

### Step 1: Understand the motion of the elevator and the ball
The elevator is moving upward with a constant acceleration of \(2 \, \text{m/s}^2\) and has a velocity of \(4 \, \text{m/s}\) at the moment the ball is projected. The ball is projected at an angle of \(30^\circ\) with a speed of \(4 \, \text{m/s}\) relative to the elevator.

### Step 2: Determine the effective acceleration acting on the ball
Since the elevator is accelerating upwards, the effective acceleration acting on the ball when it is projected will be the gravitational acceleration \(g\) plus the elevator's acceleration. Thus, the effective acceleration \(g_{\text{effective}}\) is:
\[ g_{\text{effective}} = g + a = 10 \, \text{m/s}^2 + 2 \, \text{m/s}^2 = 12 \, \text{m/s}^2 \]

### Step 3: Resolve the initial velocity of the ball into components
The initial velocity of the ball relative to the elevator is \(4 \, \text{m/s}\) at an angle of \(30^\circ\). We can resolve this into horizontal and vertical components:
- Horizontal component \(u_x = 4 \cos(30^\circ) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m/s}\)
- Vertical component \(u_y = 4 \sin(30^\circ) = 4 \times \frac{1}{2} = 2 \, \text{m/s}\)

### Step 4: Set up the equation for vertical motion
The vertical motion of the ball can be described by the equation:
\[ s_y = u_y t + \frac{1}{2} a t^2 \]
Since the ball returns to the floor of the elevator, the vertical displacement \(s_y = 0\). Therefore, we can write:
\[ 0 = 2t - \frac{1}{2} \cdot 12 t^2 \]

### Step 5: Simplify the equation
Rearranging the equation gives:
\[ 0 = 2t - 6t^2 \]
Factoring out \(t\):
\[ t(2 - 6t) = 0 \]
This gives us two solutions: \(t = 0\) (the moment of projection) and \(t = \frac{2}{6} = \frac{1}{3} \, \text{s}\).

### Step 6: Conclusion
The time taken by the ball to return to the floor of the elevator is:
\[ t = \frac{1}{3} \, \text{s} \]