A body falling freely from a given height `H` hits an inlclined plane in its path at a height `h`. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of `h//H`, the body will take the maximum time to reach the ground.

Options

A. `(1)/(3)`
B. `(1)/(2)`
C. `(2)/(5)`
D. `(2)/(3)`

Explanation

To solve the problem, we need to find the ratio \( \frac{h}{H} \) for which the body takes the maximum time to reach the ground after hitting the inclined plane. We will break down the solution into clear steps.

### Step 1: Understanding the Motion
The body falls freely from a height \( H \) and hits the inclined plane at height \( h \). After the impact, the direction of the velocity becomes horizontal.

### Step 2: Calculate Time from A to B
Let \( t_1 \) be the time taken to fall from point A (height \( H \)) to point B (height \( h \)). Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = H - h \), \( u = 0 \), and \( a = g \) (acceleration due to gravity). Therefore:
\[ H - h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ H - h = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2(H - h)}{g} \]
Thus, we find:
\[ t_1 = \sqrt{\frac{2(H - h)}{g}} \]

### Step 3: Calculate Time from B to C
Let \( t_2 \) be the time taken to fall from point B (height \( h \)) to point C (ground level). Again using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = h \), \( u = 0 \), and \( a = g \). Therefore:
\[ h = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{2h}{g} \]
Thus, we find:
\[ t_2 = \sqrt{\frac{2h}{g}} \]

### Step 4: Total Time Calculation
The total time \( T \) taken to reach the ground is:
\[ T = t_1 + t_2 = \sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}} \]
Factoring out \( \sqrt{\frac{2}{g}} \):
\[ T = \sqrt{\frac{2}{g}} \left( \sqrt{H - h} + \sqrt{h} \right) \]

### Step 5: Maximizing the Total Time
To find the value of \( h \) that maximizes \( T \), we differentiate \( T \) with respect to \( h \) and set the derivative to zero:
\[ \frac{dT}{dh} = \frac{1}{2\sqrt{H - h}}(-1) + \frac{1}{2\sqrt{h}} = 0 \]
This leads to:
\[ \frac{1}{\sqrt{H - h}} = \frac{1}{\sqrt{h}} \]
Squaring both sides gives:
\[ H - h = h \]
Rearranging gives:
\[ H = 2h \quad \Rightarrow \quad h = \frac{H}{2} \]

### Step 6: Finding the Ratio
Now, we can find the ratio \( \frac{h}{H} \):
\[ \frac{h}{H} = \frac{\frac{H}{2}}{H} = \frac{1}{2} \]

### Conclusion
The value of \( \frac{h}{H} \) for which the body takes the maximum time to reach the ground is:
\[ \frac{h}{H} = \frac{1}{2} \]