To solve the problem step by step, let's break it down into three parts as per the question.
### Given Data:
- Speed of the elevator, \( v_e = 10 \, \text{m/s} \) (upward)
- Height of the boy from the ground, \( h = 10 \, \text{m} \)
- Velocity of projection of the ball with respect to the elevator, \( v_{b/e} = 30 \, \text{m/s} \) (upward)
### Part (a): Maximum Height Attained by the Ball
1. **Calculate the velocity of the ball with respect to the ground**:
\[
v_{b/g} = v_{b/e} + v_e = 30 \, \text{m/s} + 10 \, \text{m/s} = 40 \, \text{m/s}
\]
2. **Use the formula for maximum height in projectile motion**:
The maximum height \( H \) attained by the ball can be calculated using the formula:
\[
H = \frac{v_{b/g}^2}{2g}
\]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
3. **Substituting the values**:
\[
H = \frac{(40)^2}{2 \times 10} = \frac{1600}{20} = 80 \, \text{m}
\]
4. **Calculate the total height from the ground**:
Since the ball was thrown from a height of 10 m:
\[
\text{Total height} = h + H = 10 \, \text{m} + 80 \, \text{m} = 90 \, \text{m}
\]
**Answer for Part (a)**: The maximum height attained by the ball is **90 m**.
### Part (b): Time Taken by the Ball to Meet the Elevator Again
1. **Calculate the time taken for the ball to reach its maximum height**:
The time \( t_1 \) to reach the maximum height can be calculated using:
\[
t_1 = \frac{v_{b/g}}{g} = \frac{40 \, \text{m/s}}{10 \, \text{m/s}^2} = 4 \, \text{s}
\]
2. **Calculate the total time taken to return to the same height as the elevator**:
The ball will take the same time to come down from the maximum height to the height of the elevator:
\[
t_2 = t_1 = 4 \, \text{s}
\]
3. **Total time taken to meet the elevator again**:
\[
T = t_1 + t_2 = 4 \, \text{s} + 4 \, \text{s} = 8 \, \text{s}
\]
**Answer for Part (b)**: The time taken by the ball to meet the elevator again is **8 s**.
### Part (c): Time Taken by the Ball to Reach the Ground After Crossing the Elevator
1. **Calculate the time taken to fall from the maximum height to the ground**:
The ball is at a height of 90 m when it reaches its maximum height. The time \( t_3 \) to fall from this height can be calculated using the equation of motion:
\[
h = v_{b/g} t + \frac{1}{2} g t^2
\]
Here, \( h = 90 \, \text{m} \), \( v_{b/g} = 0 \) (initial velocity at the maximum height), and \( g = 10 \, \text{m/s}^2 \).
The equation simplifies to:
\[
90 = 0 + \frac{1}{2} \cdot 10 \cdot t^2
\]
\[
90 = 5t^2 \implies t^2 = \frac{90}{5} = 18 \implies t = \sqrt{18} \approx 4.24 \, \text{s}
\]
2. **Total time taken from the moment the ball was thrown until it hits the ground**:
The total time taken from the moment the ball was thrown until it hits the ground is:
\[
T_{\text{total}} = T + t_3 = 8 \, \text{s} + 4.24 \, \text{s} \approx 12.24 \, \text{s}
\]
**Answer for Part (c)**: The time taken by the ball to reach the ground after crossing the elevator is approximately **4.24 s**.
### Summary of Answers:
- (a) Maximum height attained by the ball: **90 m**
- (b) Time taken by the ball to meet the elevator again: **8 s**
- (c) Time taken by the ball to reach the ground after crossing the elevator: **4.24 s**
