A ball is released from the top of a tower of height `H m`. After `2 s` is stopped and then instantaneously released. What will be its height after next `2 s`?.

Options

A. `(H-5) m`
B. ` (H-10) m`
C. ` (H-20) m`
D. `(H-40) m`

Explanation

To solve the problem, we need to analyze the motion of the ball in two phases:

1. **First Phase (First 2 seconds)**:
- The ball is released from the top of the tower and falls freely under gravity.
- The initial velocity (u) = 0 m/s (since it is released).
- The distance traveled by the ball in the first 2 seconds can be calculated using the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
- Here, \(g\) (acceleration due to gravity) is approximately \(10 \, \text{m/s}^2\) (for simplicity), and \(t = 2 \, \text{s}\).
- Plugging in the values:
\[ s = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 = 0 + \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{m} \]
- Therefore, after 2 seconds, the ball has fallen 20 meters.

2. **Second Phase (Next 2 seconds)**:
- After 2 seconds, the ball is stopped and then instantaneously released again from the height it reached after the first 2 seconds.
- The height from which it is released now is \(H - 20 \, \text{m}\).
- Once released again, it will fall freely for another 2 seconds.
- The distance it will fall in the next 2 seconds (again using the same equation of motion) will be the same as before:
\[ s = ut + \frac{1}{2}gt^2 = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2)^2 = 20 \, \text{m} \]
- Thus, in the second phase, it falls another 20 meters.

3. **Final Height Calculation**:
- The total distance fallen after 4 seconds is \(20 \, \text{m} + 20 \, \text{m} = 40 \, \text{m}\).
- Therefore, the height of the ball above the ground after 4 seconds is:
\[ \text{Height after 4 seconds} = H - 40 \, \text{m} \]

**Final Answer**: The height of the ball after the next 2 seconds will be \(H - 40 \, \text{m}\).

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