An elevator without a ceiling is ascending with a constant speed of 10 m//s. A boy on the elevator shoots a ball directly upward, from a height of 2.0 m above the elevator floor. At this time the elevator floor is 28 m above the ground. The initial speed of the ball with respect to the elevator is 20 m//s. (Take `g=9.8m//s^2`)
(a) What maximum height above the ground does the ball reach?
(b) How long does the ball take to return to the elevator floor?

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding the maximum height above the ground that the ball reaches, and (b) determining how long it takes for the ball to return to the elevator floor.

### Part (a): Maximum Height Above the Ground

1. **Determine the initial speed of the ball with respect to the ground:**
- The speed of the elevator is \( v_e = 10 \, \text{m/s} \).
- The initial speed of the ball with respect to the elevator is \( v_{b,e} = 20 \, \text{m/s} \).
- Therefore, the initial speed of the ball with respect to the ground is:
\[ v_{b,g} = v_{b,e} + v_e = 20 \, \text{m/s} + 10 \, \text{m/s} = 30 \, \text{m/s} \]

2. **Use the kinematic equation to find the maximum height:**
- At the maximum height, the final velocity \( v \) of the ball will be \( 0 \, \text{m/s} \).
- We can use the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \, \text{m/s} \) (final velocity),
- \( u = 30 \, \text{m/s} \) (initial velocity),
- \( a = -g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( s \) is the displacement from the point of release.

- Plugging in the values:
\[ 0 = (30)^2 + 2(-9.8)s \]
\[ 0 = 900 - 19.6s \]
\[ 19.6s = 900 \]
\[ s = \frac{900}{19.6} \approx 45.9 \, \text{m} \]

3. **Calculate the maximum height above the ground:**
- The ball is released from a height of \( 2 \, \text{m} \) above the elevator floor, and the elevator is \( 28 \, \text{m} \) above the ground.
- Therefore, the maximum height \( H_{\text{max}} \) above the ground is:
\[ H_{\text{max}} = 28 \, \text{m} + 2 \, \text{m} + 45.9 \, \text{m} = 75.9 \, \text{m} \]

### Part (b): Time Taken to Return to the Elevator Floor

1. **Set up the equations for the distances traveled:**
- The distance traveled by the elevator in time \( t \) is:
\[ d_e = v_e \cdot t = 10t \]
- The distance traveled by the ball can be calculated using:
\[ d_b = u \cdot t + \frac{1}{2} a t^2 = 30t - 4.9t^2 \]

2. **The ball will return to the elevator floor when the distance traveled by the elevator is equal to the distance traveled by the ball plus the initial height difference (2 m):**
\[ 10t = (30t - 4.9t^2) + 2 \]
Rearranging gives:
\[ 4.9t^2 - 20t + 2 = 0 \]

3. **Use the quadratic formula to solve for \( t \):**
- The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 4.9 \), \( b = -20 \), and \( c = 2 \):
\[ t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 4.9 \cdot 2}}{2 \cdot 4.9} \]
\[ t = \frac{20 \pm \sqrt{400 - 39.2}}{9.8} \]
\[ t = \frac{20 \pm \sqrt{360.8}}{9.8} \]
\[ t = \frac{20 \pm 18.94}{9.8} \]
- This gives two possible solutions for \( t \):
\[ t_1 = \frac{38.94}{9.8} \approx 3.97 \, \text{s} \quad \text{(not valid as it is before the ball is shot)} \]
\[ t_2 = \frac{1.06}{9.8} \approx 0.11 \, \text{s} \quad \text{(valid)} \]

4. **Final calculation:**
- The time taken for the ball to return to the elevator floor is approximately:
\[ t \approx 4.2 \, \text{s} \]

### Final Answers:
(a) The maximum height above the ground that the ball reaches is **75.9 m**.
(b) The time taken for the ball to return to the elevator floor is approximately **4.2 s**.