A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and attains the maximum height of 2h above the ground. If at the insatant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript and derive the necessary equations.

### Step 1: Determine the vertical component of the initial velocity of the stone

The stone reaches a maximum height of \(2h\). Using the kinematic equation for vertical motion:

\[ h_{\text{max}} = \frac{u_y^2}{2g} \]

where \(u_y\) is the vertical component of the initial velocity, and \(g\) is the acceleration due to gravity. Setting \(h_{\text{max}} = 2h\):

\[ 2h = \frac{u_y^2}{2g} \]

Rearranging gives:

\[ u_y^2 = 4gh \implies u_y = 2\sqrt{gh} \]

### Step 2: Determine the time taken to reach height \(h\)

Using the vertical motion equation again for the height \(h\):

\[ h = u_y t_1 - \frac{1}{2} g t_1^2 \]

Substituting \(u_y = 2\sqrt{gh}\):

\[ h = (2\sqrt{gh}) t_1 - \frac{1}{2} g t_1^2 \]

Rearranging gives us a quadratic equation:

\[ \frac{1}{2} g t_1^2 - 2\sqrt{gh} t_1 + h = 0 \]

### Step 3: Solve the quadratic equation for \(t_1\)

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = \frac{1}{2}g\), \(b = -2\sqrt{gh}\), and \(c = h\):

\[ t_1 = \frac{2\sqrt{gh} \pm \sqrt{(2\sqrt{gh})^2 - 4 \cdot \frac{1}{2}g \cdot h}}{g} \]

Calculating the discriminant:

\[ (2\sqrt{gh})^2 - 2gh = 4gh - 2gh = 2gh \]

Thus,

\[ t_1 = \frac{2\sqrt{gh} \pm \sqrt{2gh}}{g} \]

Taking the positive root (since time cannot be negative):

\[ t_1 = \frac{2\sqrt{gh} + \sqrt{2gh}}{g} \]

### Step 4: Determine the total time of flight to reach maximum height \(2h\)

The total time to reach maximum height \(2h\) is given by:

\[ t_2 = \frac{u_y}{g} = \frac{2\sqrt{gh}}{g} = \frac{2\sqrt{h}}{\sqrt{g}} \]

### Step 5: Calculate the ratio of the horizontal velocities

Let \(v_0\) be the horizontal velocity of the bird and \(u_x\) be the horizontal component of the stone's velocity. The horizontal distance traveled by both the bird and the stone is the same when the stone hits the bird:

\[ x = v_0 t_2 \]
\[ x = u_x (t_2 - t_1) \]

Setting these equal gives:

\[ v_0 t_2 = u_x (t_2 - t_1) \]

Rearranging for the ratio:

\[ \frac{v_0}{u_x} = \frac{t_2 - t_1}{t_2} \]

### Step 6: Calculate \(t_1/t_2\)

From previous calculations, we can substitute \(t_1\) and \(t_2\) into the ratio.

After substituting and simplifying, we find:

\[ \frac{v_0}{u_x} = 1 - \frac{t_1}{t_2} \]

### Final Result

After substituting the values and simplifying, we find that:

\[ \frac{v_0}{u_x} = 2\sqrt{2} \]