An elevator without a ceiling is ascending up with an acceleration of `5 ms^-2.` A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of `10 ms^-1` and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is `15 ms^-1` with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. (`g=10 ms^-2`)
1. The time in which the ball strikes the floor of elevator is given by

Options

A. 2.13 s
B. 2.0 s
C. 1.0 s
D. 3.12 s

Explanation

To solve the problem, we need to analyze the motion of the ball with respect to the ground and the elevator.

### Step 1: Determine the initial conditions
- The ball is shot from a height of 2 m above the floor of the elevator.
- The initial speed of the ball with respect to the elevator is 15 m/s upward.
- The elevator is moving upward with a speed of 10 m/s.
- The acceleration of the elevator is 5 m/s² upward.
- The acceleration due to gravity (g) is 10 m/s² downward.

### Step 2: Calculate the initial speed of the ball with respect to the ground
The initial speed of the ball with respect to the ground can be calculated by adding the speed of the elevator to the speed of the ball with respect to the elevator:
\[ v_{b,0} = v_{e} + v_{b/e} \]
Where:
- \( v_{e} = 10 \, \text{m/s} \) (speed of the elevator)
- \( v_{b/e} = 15 \, \text{m/s} \) (speed of the ball with respect to the elevator)

Substituting the values:
\[ v_{b,0} = 10 \, \text{m/s} + 15 \, \text{m/s} = 25 \, \text{m/s} \]

### Step 3: Set up the equation of motion for the ball
The ball is subject to gravity, so we can use the following kinematic equation to find the time \( t \) when the ball hits the floor of the elevator:
\[ s = v_{b,0} t - \frac{1}{2} g t^2 \]
Where:
- \( s = -2 \, \text{m} \) (the ball falls 2 m to reach the floor of the elevator)
- \( g = 10 \, \text{m/s}^2 \)

Substituting the values:
\[ -2 = 25t - \frac{1}{2} \cdot 10 t^2 \]
This simplifies to:
\[ -2 = 25t - 5t^2 \]
Rearranging gives us a quadratic equation:
\[ 5t^2 - 25t - 2 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 5 \)
- \( b = -25 \)
- \( c = -2 \)

Calculating the discriminant:
\[ b^2 - 4ac = (-25)^2 - 4 \cdot 5 \cdot (-2) = 625 + 40 = 665 \]

Now substituting into the quadratic formula:
\[ t = \frac{25 \pm \sqrt{665}}{10} \]

Calculating \( \sqrt{665} \) gives approximately 25.77:
\[ t = \frac{25 \pm 25.77}{10} \]

This results in two possible solutions:
1. \( t = \frac{50.77}{10} = 5.077 \, \text{s} \) (valid solution)
2. \( t = \frac{-0.77}{10} \) (not valid as time cannot be negative)

### Final Answer
The time in which the ball strikes the floor of the elevator is approximately:
\[ t \approx 5.08 \, \text{s} \]