An open elevator is ascending with constant speed `v=10m//s.` A ball is thrown vertically up by a boy on the lift when he is at a height `h=10m` from the ground. The velocity of projection is `v=30 m//s` with respect to elevator. Find
(a) the maximum height attained by the ball.
(b) the time taken by the ball to meet the elevator again.
(c) time taken by the ball to reach the ground after crossing the elevator.

Explanation

To solve the problem step by step, we will break it down into three parts as per the question.

### Given Data:
- Speed of the elevator, \( v_{lift} = 10 \, \text{m/s} \)
- Height of the boy from the ground, \( h = 10 \, \text{m} \)
- Velocity of projection of the ball with respect to the elevator, \( v_{projection} = 30 \, \text{m/s} \)

### Step 1: Find the velocity of the ball with respect to the ground.
The velocity of the ball with respect to the ground is the sum of the velocity of the elevator and the velocity of projection of the ball:

\[ v_{ball} = v_{projection} + v_{lift} = 30 \, \text{m/s} + 10 \, \text{m/s} = 40 \, \text{m/s} \]

### Step 2: Calculate the maximum height attained by the ball.
The maximum height attained by the ball can be calculated using the formula for maximum height in projectile motion:

\[ h_{max} = h + \frac{v_{ball}^2}{2g} \]

Where:
- \( h \) is the initial height (10 m)
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))

Substituting the values:

\[ h_{max} = 10 \, \text{m} + \frac{(40 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} \]
\[ h_{max} = 10 \, \text{m} + \frac{1600}{20} \]
\[ h_{max} = 10 \, \text{m} + 80 \, \text{m} = 90 \, \text{m} \]

### Step 3: Find the time taken by the ball to meet the elevator again.
The distance traveled by the elevator in time \( t \) is given by:

\[ s_{lift} = v_{lift} \cdot t = 10t \]

The displacement of the ball can be calculated using the equation of motion:

\[ s_{ball} = v_{ball} \cdot t - \frac{1}{2} g t^2 = 40t - 5t^2 \]

At the time when the ball meets the elevator again, the distances traveled by both will be equal:

\[ 10t = 40t - 5t^2 \]

Rearranging gives:

\[ 5t^2 - 30t = 0 \]
\[ t(5t - 30) = 0 \]

Thus, \( t = 0 \) (initial condition) or \( t = 6 \, \text{s} \).

### Step 4: Calculate the time taken by the ball to reach the ground after crossing the elevator.
The ball is thrown from a height of 10 m and will fall to the ground. The total displacement when it reaches the ground is -10 m. Using the equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = -10 \, \text{m} \)
- \( u = 40 \, \text{m/s} \) (upward)
- \( a = -10 \, \text{m/s}^2 \) (downward)

Substituting the values:

\[ -10 = 40t - 5t^2 \]
\[ 5t^2 - 40t - 10 = 0 \]

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ t = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 5 \cdot (-10)}}{2 \cdot 5} \]
\[ t = \frac{40 \pm \sqrt{1600 + 200}}{10} = \frac{40 \pm \sqrt{1800}}{10} \]
\[ t = \frac{40 \pm 42.426}{10} \]

Calculating the two possible values gives:

1. \( t = \frac{82.426}{10} = 8.2426 \, \text{s} \)
2. \( t = \frac{-2.426}{10} \) (not physically meaningful)

The time taken by the ball to reach the ground after crossing the elevator is:

\[ t_{ground} = 8.2426 - 6 = 2.2426 \, \text{s} \approx 2.24 \, \text{s} \]

### Summary of Results:
(a) Maximum height attained by the ball: **90 m**
(b) Time taken by the ball to meet the elevator again: **6 s**
(c) Time taken by the ball to reach the ground after crossing the elevator: **2.24 s**