An open lift is moving upward with velocity `10 m//s.` It has an upward acceleration of `2 m//s^2.` A ball is projected upwards with velocity `20 m//s` relative to ground. Find
(a) time when ball again meets the lift
(b) displacement of lift and ball at that instant.
(c) distance travelled by the ball upto that instant.
Take `g=10 m//s^2`

Explanation

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as requested.

### Given Data:
- Velocity of the lift, \( v_L = 10 \, \text{m/s} \) (upward)
- Acceleration of the lift, \( a_L = 2 \, \text{m/s}^2 \) (upward)
- Initial velocity of the ball, \( u_B = 20 \, \text{m/s} \) (upward)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (downward)

### (a) Time when the ball again meets the lift

1. **Displacement of the Lift**:
The displacement \( S_L \) of the lift after time \( t \) can be calculated using the equation of motion:
\[ S_L = v_L t + \frac{1}{2} a_L t^2 \]
Substituting the values:
\[ S_L = 10t + \frac{1}{2} \cdot 2 \cdot t^2 = 10t + t^2 \]

2. **Displacement of the Ball**:
The displacement \( S_B \) of the ball after time \( t \) is given by:
\[ S_B = u_B t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ S_B = 20t - \frac{1}{2} \cdot 10 \cdot t^2 = 20t - 5t^2 \]

3. **Setting the Displacements Equal**:
Since we want to find the time when the ball meets the lift again, we set \( S_L = S_B \):
\[ 10t + t^2 = 20t - 5t^2 \]
Rearranging gives:
\[ 6t^2 - 10t = 0 \]
Factoring out \( t \):
\[ t(6t - 10) = 0 \]
Thus, \( t = 0 \) (initial position) or \( 6t - 10 = 0 \) leading to:
\[ t = \frac{10}{6} = \frac{5}{3} \, \text{seconds} \]

### (b) Displacement of the lift and ball at that instant

1. **Using the time \( t = \frac{5}{3} \) seconds**:
Substitute \( t \) into the displacement equation for the lift:
\[ S_L = 10 \left(\frac{5}{3}\right) + \left(\frac{5}{3}\right)^2 \]
Calculating:
\[ S_L = \frac{50}{3} + \frac{25}{9} = \frac{150}{9} + \frac{25}{9} = \frac{175}{9} \, \text{meters} \]

### (c) Distance travelled by the ball up to that instant

1. **Finding the time when the ball's velocity becomes zero**:
The time \( t_0 \) when the ball's velocity becomes zero can be found using:
\[ v = u_B - g t \implies 0 = 20 - 10t \implies t = 2 \, \text{seconds} \]

2. **Since \( \frac{5}{3} < 2 \)**:
The ball is still moving upwards at \( t = \frac{5}{3} \) seconds, so the distance travelled is equal to the displacement:
\[ S_B = 20 \left(\frac{5}{3}\right) - 5 \left(\frac{5}{3}\right)^2 \]
Calculating:
\[ S_B = \frac{100}{3} - 5 \cdot \frac{25}{9} = \frac{100}{3} - \frac{125}{9} = \frac{300}{9} - \frac{125}{9} = \frac{175}{9} \, \text{meters} \]

### Summary of Results:
- (a) Time when the ball meets the lift again: \( t = \frac{5}{3} \, \text{seconds} \)
- (b) Displacement of the lift and ball at that instant: \( S_L = S_B = \frac{175}{9} \, \text{meters} \approx 19.4 \, \text{meters} \)
- (c) Distance travelled by the ball up to that instant: \( S_B = \frac{175}{9} \, \text{meters} \approx 19.4 \, \text{meters} \)