A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with at speed of `20 m//s`. Two seconds later, an identical stone is thrown vertically downwards with the same speed of `20 m`,. Then

Options

A. the relative velocity between the two stones remain, constant till one hits the ground
B. both will have the same kinetic energy, when they hit the ground
C. the time interval between their hitting the ground it `2s`
D. if the collision on the ground is perfectly elastic, both will rise to the same height above the ground

Explanation

To solve the problem step by step, we will analyze the motion of the two stones thrown by the man standing on the terrace.

### Step 1: Analyze the first stone (thrown upwards)
- The first stone is thrown vertically upwards with an initial speed \( u_1 = 20 \, \text{m/s} \).
- The acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \) acts downwards.
- The time taken to reach the maximum height can be calculated using the formula:
\[ v = u + at \]
where \( v = 0 \) at the maximum height.
\[ 0 = 20 - 9.81t \implies t = \frac{20}{9.81} \approx 2.04 \, \text{s} \]
- The total time of flight until it hits the ground is the time to reach maximum height plus the time to fall back down. The time to fall back down will be the same as the time taken to go up, so:
\[ \text{Total time for first stone} = 2.04 + 2.04 \approx 4.08 \, \text{s} \]

### Step 2: Analyze the second stone (thrown downwards)
- The second stone is thrown downwards with the same initial speed \( u_2 = 20 \, \text{m/s} \) but 2 seconds later.
- The time taken for the second stone to hit the ground can be calculated similarly. It will fall for the entire duration of its flight until it hits the ground. Since it is thrown 2 seconds later, it will have:
\[ \text{Total time for second stone} = 4.08 - 2 \approx 2.08 \, \text{s} \]

### Step 3: Relative velocity and acceleration
- Both stones experience the same gravitational acceleration \( g \) downwards.
- The relative acceleration between the two stones is zero because both are subjected to the same gravitational force.
- Since the relative acceleration is zero, the relative velocity remains constant throughout their motion.

### Step 4: Kinetic energy at the point of impact
- The kinetic energy of both stones when they hit the ground can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
- Both stones will have the same speed just before hitting the ground due to the symmetry of their motions (upwards and downwards) and the same initial speed. Therefore, their kinetic energies will also be equal.

### Step 5: Time interval between hitting the ground
- The first stone takes approximately 4.08 seconds to hit the ground, while the second stone takes approximately 2.08 seconds.
- The time interval between their impacts is:
\[ \text{Time interval} = 4.08 - 2.08 = 2 \, \text{s} \]

### Step 6: Height reached after collision (if perfectly elastic)
- If the collision is perfectly elastic, both stones will rebound to the same height because they have the same speed just before hitting the ground, and the coefficient of restitution is equal.

### Conclusion
- The statements given in the question can be verified as follows:
1. The relative velocity between the two stones remains constant until one hits the ground. **True**
2. Both will have the same kinetic energy when they hit the ground. **True**
3. The time interval between their hitting the ground is 2 seconds. **True**
4. If the collision on the ground is perfectly elastic, both will rise to the same height above the ground. **True**