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.

Explanation

To solve the problem, we need to determine the value of \( \frac{h}{H} \) for which the body takes the maximum time to reach the ground after hitting the inclined plane. Here’s a step-by-step solution:

### Step 1: Understand the Problem
A body is falling freely from a height \( H \) and hits an inclined plane at a height \( h \). After the impact, the body moves horizontally. We need to find the ratio \( \frac{h}{H} \) that maximizes the time taken to reach the ground.

### Step 2: Calculate the Time to Hit the Inclined Plane
The body falls a distance of \( H - h \) before hitting the inclined plane. Using the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( s = H - h \), \( u = 0 \) (initial velocity), and \( g \) is the acceleration due to gravity, we can find the time \( t_0 \) taken to fall this distance:
\[ H - h = 0 \cdot t_0 + \frac{1}{2} g t_0^2 \implies t_0 = \sqrt{\frac{2(H - h)}{g}} \]

### Step 3: Calculate the Velocity at Impact
Using the third equation of motion, we can find the velocity \( v \) just before hitting the inclined plane:
\[ v^2 = u^2 + 2g(H - h) \implies v = \sqrt{2g(H - h)} \]

### Step 4: Calculate the Time to Fall from Height \( h \)
After hitting the inclined plane, the body moves horizontally with velocity \( v \) and falls a vertical distance \( h \). The time of flight \( t_f \) to fall this height is given by:
\[ h = \frac{1}{2} g t_f^2 \implies t_f = \sqrt{\frac{2h}{g}} \]

### Step 5: Total Time Calculation
The total time \( T \) taken to reach the ground is the sum of the time to hit the inclined plane and the time to fall from height \( h \):
\[ T = t_0 + t_f = \sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}} \]
Factoring out \( \sqrt{\frac{2}{g}} \):
\[ T = \sqrt{\frac{2}{g}} \left( \sqrt{H - h} + \sqrt{h} \right) \]

### Step 6: Maximizing the Total Time
To find the value of \( h \) that maximizes \( T \), we can differentiate \( T \) with respect to \( h \) and set the derivative to zero:
\[ \frac{dT}{dh} = \frac{1}{2\sqrt{H - h}}(-1) + \frac{1}{2\sqrt{h}}(1) = 0 \]
This leads to:
\[ \sqrt{h} = \sqrt{H - h} \]
Squaring both sides gives:
\[ h = H - h \implies 2h = H \implies h = \frac{H}{2} \]

### Step 7: Finding the Ratio
Thus, the ratio \( \frac{h}{H} \) is:
\[ \frac{h}{H} = \frac{\frac{H}{2}}{H} = \frac{1}{2} \]

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