A platform is moving upwards with a constant acceleration of `2 m s^-2`. At time `t = 0`, a boy standing on the platform throws a ball upwards with a relative speed of `8 m s^-1`. At this instant, platform was at the height of `4 m` from the ground and was moving with a speed of `2 m s^-1`. Take `g = 0 m s^-2`. Find
(a) when and where the ball strikes the platform.
(b) the maximum height attained by the ball from the ground.
( c) the maximum distance of the ball from the platform.

Explanation

To solve the problem step by step, let's break it down into parts (a), (b), and (c) as required.

### Given Data:
- Acceleration of the platform, \( a_p = 2 \, \text{m/s}^2 \)
- Initial velocity of the platform, \( u_p = 2 \, \text{m/s} \)
- Relative speed of the ball thrown upwards, \( u_b = 8 \, \text{m/s} \)
- Initial height of the platform, \( h_p = 4 \, \text{m} \)
- Acceleration due to gravity, \( g = 0 \, \text{m/s}^2 \) (as per the question)

### (a) When and where the ball strikes the platform

1. **Calculate the actual initial velocity of the ball:**
\[ u = u_b + u_p = 8 \, \text{m/s} + 2 \, \text{m/s} = 10 \, \text{m/s} \]

2. **Calculate the relative acceleration of the ball with respect to the platform:**
\[ a_{relative} = -g - a_p = -0 - 2 = -2 \, \text{m/s}^2 \]

3. **Use the equation of motion to find the time \( t \) when the ball strikes the platform:**
The relative displacement \( s_{relative} \) when the ball strikes the platform is 0:
\[ s_{relative} = u_b t + \frac{1}{2} a_{relative} t^2 = 0 \]
\[ 0 = 8t - \frac{1}{2} \cdot 2t^2 \]
\[ 0 = 8t - t^2 \]
\[ t(t - 8) = 0 \]
Thus, \( t = 0 \) or \( t = 8 \, \text{s} \).

4. **Calculate the height of the platform when the ball strikes it:**
Use the equation of motion for the platform:
\[ h = h_p + u_p t + \frac{1}{2} a_p t^2 \]
\[ h = 4 + 2 \cdot 8 + \frac{1}{2} \cdot 2 \cdot (8^2) \]
\[ h = 4 + 16 + 64 = 84 \, \text{m} \]

### (b) The maximum height attained by the ball from the ground

1. **Use the equation of motion to find the maximum height:**
At maximum height, the final velocity \( v = 0 \):
\[ v^2 = u^2 + 2as \]
\[ 0 = (10)^2 + 2(-2)s \]
\[ 0 = 100 - 4s \]
\[ 4s = 100 \implies s = 25 \, \text{m} \]

2. **Calculate the total maximum height from the ground:**
\[ \text{Total height} = h_p + s = 4 + 25 = 29 \, \text{m} \]

### (c) The maximum distance of the ball from the platform

1. **Calculate the maximum distance of the ball from the platform:**
The maximum distance occurs when the ball reaches its maximum height. The height of the platform at that time is:
\[ h_{platform} = h_p + u_p t + \frac{1}{2} a_p t^2 \]
where \( t = 8 \, \text{s} \):
\[ h_{platform} = 4 + 2 \cdot 8 + \frac{1}{2} \cdot 2 \cdot (8^2) = 84 \, \text{m} \]

2. **Calculate the maximum distance:**
\[ \text{Distance} = \text{Maximum height of the ball} - \text{Height of the platform} \]
\[ \text{Distance} = 29 - 84 = -55 \, \text{m} \]
This indicates that the ball is below the platform at its maximum height.

### Summary of Results:
- (a) The ball strikes the platform at \( t = 8 \, \text{s} \) and at a height of \( 84 \, \text{m} \).
- (b) The maximum height attained by the ball from the ground is \( 29 \, \text{m} \).
- (c) The maximum distance of the ball from the platform is \( -55 \, \text{m} \) (indicating the ball is below the platform).