A particle is projected vertically upwards from the earth. It crosses the same point at t=2 sec and t=8 sec. Find out `[g=10 ms^(-2)]`
(i) the time for which particle remains in air
(ii) intial velocity of particle
(iii) maximum height attained by the particle
(iv) height of the particle at t=2 sec and t=8 sec from earth

Answer

To solve the problem step by step, we will break down each part of the question and use the equations of motion for uniformly accelerated motion.

### Given Data:
- The particle crosses the same point at \( t = 2 \, \text{s} \) and \( t = 8 \, \text{s} \).
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \).

### (i) Time for which the particle remains in air

1. **Understanding the motion**: The particle is projected upwards, reaches a maximum height, and then returns back down. The total time of flight can be calculated as the time taken to reach the maximum height and the time taken to return back to the same height.

2. **Finding the time of flight**: The particle crosses the same point at \( t = 2 \, \text{s} \) on the way up and at \( t = 8 \, \text{s} \) on the way down. The total time of flight \( T \) is given by:
\[ T = t_{\text{up}} + t_{\text{down}} = 8 \, \text{s} \]
Therefore, the total time for which the particle remains in air is:
\[ T = 8 \, \text{s} - 0 \, \text{s} = 8 \, \text{s} \]

### (ii) Initial velocity of the particle

1. **Using the time to reach the maximum height**: The time to reach the maximum height is half of the total time of flight:
\[ t_{\text{up}} = \frac{T}{2} = \frac{8}{2} = 4 \, \text{s} \]

2. **Using the equation of motion**: The final velocity at the maximum height is \( 0 \, \text{m/s} \). We can use the equation:
\[ v = u + at \]
where \( v = 0 \), \( a = -g = -10 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \):
\[ 0 = u - 10 \times 4 \]
\[ u = 40 \, \text{m/s} \]

### (iii) Maximum height attained by the particle

1. **Using the equation of motion**: We can use the equation:
\[ v^2 = u^2 + 2as \]
where \( v = 0 \), \( u = 40 \, \text{m/s} \), and \( a = -10 \, \text{m/s}^2 \):
\[ 0 = (40)^2 + 2(-10)s \]
\[ 0 = 1600 - 20s \]
\[ 20s = 1600 \implies s = \frac{1600}{20} = 80 \, \text{m} \]

### (iv) Height of the particle at \( t = 2 \, \text{s} \) and \( t = 8 \, \text{s} \)

1. **Using the equation of motion**: The height at any time \( t \) can be calculated using:
\[ s = ut + \frac{1}{2}at^2 \]
For \( t = 2 \, \text{s} \):
\[ s = 40 \times 2 + \frac{1}{2}(-10)(2^2) \]
\[ s = 80 - 20 = 60 \, \text{m} \]

2. **For \( t = 8 \, \text{s} \)**:
\[ s = 40 \times 8 + \frac{1}{2}(-10)(8^2) \]
\[ s = 320 - 320 = 0 \, \text{m} \]

### Summary of Answers:
1. Time in air: \( 8 \, \text{s} \)
2. Initial velocity: \( 40 \, \text{m/s} \)
3. Maximum height: \( 80 \, \text{m} \)
4. Height at \( t = 2 \, \text{s} \): \( 60 \, \text{m} \) and at \( t = 8 \, \text{s} \): \( 0 \, \text{m} \)