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 form the floor of the lift with a speed of `4m//s` relative to the floor at an elevation of `30^(@)`. Time taken by the ball to return the floor is `(g=10ms^(2))`

Options

A. (a)0.5 s
B. (b)0.33 s
C. (c)0.25 s
D. (d)1 s

Explanation

To solve the problem step by step, we will analyze the motion of the ball projected from the elevator.

### Step 1: Identify the components of the initial velocity of the ball.
The ball is projected with a speed of \(4 \, \text{m/s}\) at an angle of \(30^\circ\) relative to the floor of the elevator.

- The horizontal component of the velocity (\(u_x\)):
\[ u_x = 4 \cos(30^\circ) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m/s} \]

- The vertical component of the velocity (\(u_y\)):
\[ u_y = 4 \sin(30^\circ) = 4 \times \frac{1}{2} = 2 \, \text{m/s} \]

### Step 2: Determine the effective acceleration acting on the ball.
The elevator is accelerating upwards with \(a = 2 \, \text{m/s}^2\) and the acceleration due to gravity is \(g = 10 \, \text{m/s}^2\). Therefore, the effective acceleration (\(a_{\text{eff}}\)) acting on the ball is:
\[ a_{\text{eff}} = g + a = 10 + 2 = 12 \, \text{m/s}^2 \]

### Step 3: Calculate the time of flight until the ball returns to the floor of the elevator.
The time of flight can be calculated using the formula:
\[ t = \frac{2u_y}{a_{\text{eff}}} \]
Substituting the values:
\[ t = \frac{2 \times 2}{12} = \frac{4}{12} = \frac{1}{3} \, \text{s} \approx 0.33 \, \text{s} \]

### Conclusion
The time taken by the ball to return to the floor of the elevator is approximately \(0.33 \, \text{s}\).