A ball is dropped from a tower of height `h` under gravity. If it takes `4s` to reach the ground from height `h/2`, then time taken by it to reach from `h` to `h/2` is nearly:

Options

A. `9.65s`
B. `6.35s`
C. `8.35s`
D. `5.65s`

Answer

To solve the problem, we need to find the time taken by the ball to fall from height \( h \) to height \( \frac{h}{2} \) given that it takes \( 4 \) seconds to fall from height \( \frac{h}{2} \) to the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is dropped from a height \( h \).
- It takes \( 4 \) seconds to fall from \( \frac{h}{2} \) to the ground.

2. **Using the Kinematic Equation**:
- The distance fallen under gravity can be described by the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance fallen, \( u \) is the initial velocity (which is \( 0 \) since the ball is dropped), \( a \) is the acceleration due to gravity (\( g \)), and \( t \) is the time taken.

3. **Distance from \( \frac{h}{2} \) to Ground**:
- For the distance from \( \frac{h}{2} \) to the ground, we have:
\[ \frac{h}{2} = 0 \cdot 4 + \frac{1}{2} g (4^2) \]
- Simplifying this gives:
\[ \frac{h}{2} = \frac{1}{2} g \cdot 16 \]
\[ h = 16g \]

4. **Finding Total Time to Fall from \( h \) to Ground**:
- Now, we need to find the total time \( T \) taken to fall from height \( h \) to the ground:
\[ h = 0 \cdot T + \frac{1}{2} g T^2 \]
- Substituting \( h = 16g \):
\[ 16g = \frac{1}{2} g T^2 \]
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 16 = \frac{1}{2} T^2 \]
- Multiplying both sides by \( 2 \):
\[ 32 = T^2 \]
- Taking the square root:
\[ T = \sqrt{32} = 4\sqrt{2} \approx 5.66 \text{ seconds} \]

5. **Finding Time from \( h \) to \( \frac{h}{2} \)**:
- The time taken to fall from \( h \) to \( \frac{h}{2} \) is:
\[ T - 4 \text{ seconds} \]
- Therefore:
\[ \text{Time} = 5.66 - 4 = 1.66 \text{ seconds} \]

### Final Answer:
The time taken by the ball to reach from height \( h \) to height \( \frac{h}{2} \) is approximately \( 1.66 \) seconds.

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