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 and the elevator separately and then relate their displacements to find the time when the ball strikes the floor of the elevator.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The elevator is moving upward with an initial velocity \( v_e = 10 \, \text{m/s} \) and an upward acceleration \( a_e = 5 \, \text{m/s}^2 \).
- The ball is shot upward from a height of \( h = 2 \, \text{m} \) above the floor of the elevator with an initial speed \( u_b = 15 \, \text{m/s} \) relative to the elevator.

2. **Calculate the Initial Velocity of the Ball:**
- The initial velocity of the ball with respect to the ground is given by:
\[ u = u_b + v_e = 15 \, \text{m/s} + 10 \, \text{m/s} = 25 \, \text{m/s} \]

3. **Set Up the Equations of Motion:**
- The ball's motion can be described by the equation:
\[ s_b = ut - \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
- The displacement of the ball from its initial position (2 m above the elevator floor) is:
\[ s_b = 2 + 25t - 5t^2 \]

4. **Calculate the Displacement of the Elevator:**
- The displacement of the elevator can be described by the equation:
\[ s_e = v_e t + \frac{1}{2} a_e t^2 = 10t + \frac{1}{2} \cdot 5 \cdot t^2 = 10t + 2.5t^2 \]

5. **Set Up the Condition for When the Ball Strikes the Floor:**
- The ball will strike the floor of the elevator when the displacement of the ball equals the displacement of the elevator plus the height of the ball above the floor of the elevator (2 m):
\[ s_b = s_e + 2 \]
- Thus, we have:
\[ 2 + 25t - 5t^2 = 10t + 2.5t^2 + 2 \]

6. **Simplify the Equation:**
- Canceling the 2 on both sides:
\[ 25t - 5t^2 = 10t + 2.5t^2 \]
- Rearranging gives:
\[ 25t - 10t - 5t^2 - 2.5t^2 = 0 \]
\[ 15t - 7.5t^2 = 0 \]
- Factoring out \( t \):
\[ t(15 - 7.5t) = 0 \]

7. **Solve for Time:**
- The solutions are:
\[ t = 0 \quad \text{or} \quad 15 - 7.5t = 0 \implies t = 2 \, \text{s} \]

8. **Final Calculation:**
- Since \( t = 0 \) corresponds to the moment the ball is shot, we take \( t = 2 \) seconds as the time when the ball strikes the floor of the elevator.

### Conclusion:
The time in which the ball strikes the floor of the elevator is \( t = 2 \, \text{s} \).

---