A particle is dropped from the top a tower `h` metre high and at the same moment another particle is projected upward from the bottom. They meet the upper one has descended a distance `h//n` . Show that thevelocities of the two when they meet are in the ratio `2`: `(n-2)` and that the initial velocity of the particle projected upis `sqrt((1//2))n gh`.

Explanation

To solve the problem, we will analyze the motion of both particles step by step.

### Step 1: Analyze the motion of the first particle (dropped from height h)

The first particle is dropped from a height \( h \). The distance it descends when they meet is given as \( \frac{h}{n} \).

Using the equation of motion for free fall:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( s \) is the distance fallen, \( u \) is the initial velocity (which is 0 for a dropped object), \( g \) is the acceleration due to gravity, and \( t \) is the time of fall.

Substituting the known values:
\[ \frac{h}{n} = 0 + \frac{1}{2}gt^2 \]
This simplifies to:
\[ \frac{h}{n} = \frac{1}{2}gt^2 \]
From this, we can express \( t^2 \):
\[ t^2 = \frac{2h}{gn} \]

### Step 2: Calculate the velocity of the first particle when they meet

The velocity of the first particle when it has fallen a distance of \( \frac{h}{n} \) can be calculated using:
\[ v = u + gt \]
Since \( u = 0 \):
\[ v_1 = gt \]
Substituting \( t \) from the previous step:
\[ v_1 = g \sqrt{\frac{2h}{gn}} = \sqrt{2gh} \]

### Step 3: Analyze the motion of the second particle (projected upwards)

Let the initial velocity of the second particle (projected upwards) be \( u \). The distance it travels when they meet is:
\[ h - \frac{h}{n} = h\left(1 - \frac{1}{n}\right) = h\frac{n-1}{n} \]

Using the equation of motion for the upward motion:
\[ s = ut - \frac{1}{2}gt^2 \]
Substituting the known values:
\[ h\frac{n-1}{n} = ut - \frac{1}{2}gt^2 \]

### Step 4: Substitute \( t^2 \) into the equation for the second particle

From Step 1, we know:
\[ t^2 = \frac{2h}{gn} \]
Thus,
\[ \frac{1}{2}gt^2 = \frac{1}{2}g\left(\frac{2h}{gn}\right) = \frac{h}{n} \]

Now substituting this back into the equation for the second particle:
\[ h\frac{n-1}{n} = ut - \frac{h}{n} \]
Rearranging gives:
\[ h\frac{n-1}{n} + \frac{h}{n} = ut \]
\[ h\frac{n}{n} = ut \]
\[ h = ut \]
Thus,
\[ u = \frac{h}{t} \]

### Step 5: Substitute \( t \) to find \( u \)

Substituting \( t = \sqrt{\frac{2h}{gn}} \):
\[ u = \frac{h}{\sqrt{\frac{2h}{gn}}} = \sqrt{\frac{hgn}{2}} \]

### Step 6: Find the velocities ratio

Now, we have:
- \( v_1 = \sqrt{2gh} \)
- \( v_2 = u - gt = \sqrt{\frac{hgn}{2}} - g\sqrt{\frac{2h}{gn}} \)

Calculating \( v_2 \):
\[ v_2 = \sqrt{\frac{hgn}{2}} - \sqrt{2gh} = \sqrt{2gh}\left(\frac{n-2}{n}\right) \]

### Step 7: Ratio of velocities

The ratio of the velocities \( v_1 : v_2 \) becomes:
\[ \frac{v_1}{v_2} = \frac{\sqrt{2gh}}{\sqrt{2gh}\left(\frac{n-2}{n}\right)} = \frac{n}{n-2} \]
Thus, the ratio of the velocities is:
\[ v_1 : v_2 = 2 : (n - 2) \]

### Step 8: Conclusion for initial velocity

The initial velocity of the particle projected upwards is:
\[ u = \sqrt{\frac{1}{2}} n \sqrt{gh} \]