A particle of mass `2m` is projected at an angle of `45^@` with horizontal with a velocity of `20sqrt2m//s`. After `1s` explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take `g=10m//s^2`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial velocity components
The particle of mass \(2m\) is projected at an angle of \(45^\circ\) with a velocity of \(20\sqrt{2} \, \text{m/s}\). We can find the horizontal and vertical components of the initial velocity using trigonometric functions.

\[ v_{0x} = v_0 \cos(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

\[ v_{0y} = v_0 \sin(45^\circ) = 20\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 20 \, \text{m/s} \]

### Step 2: Calculate the vertical position after 1 second
Next, we need to find the vertical position of the particle after 1 second. The vertical displacement can be calculated using the equation of motion:

\[ y = v_{0y} t - \frac{1}{2} g t^2 \]

Substituting the values:

\[ y = 20 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 20 - 5 = 15 \, \text{m} \]

### Step 3: Determine the velocity just before the explosion
We need to find the vertical velocity of the particle just before the explosion occurs. This can be calculated using the equation:

\[ v_{y} = v_{0y} - g t \]

Substituting the values:

\[ v_{y} = 20 - 10 \cdot 1 = 10 \, \text{m/s} \]

### Step 4: Analyze the explosion
At the moment of the explosion, the particle splits into two equal masses, each of mass \(m\). One part comes to rest, while the other part will have the same horizontal velocity and the vertical velocity just before the explosion.

The horizontal velocity of the second part remains:

\[ v_{x} = v_{0x} = 20 \, \text{m/s} \]

The vertical velocity of the second part just after the explosion is:

\[ v_{y} = 10 \, \text{m/s} \]

### Step 5: Calculate the resultant velocity of the second part
The resultant velocity of the second part can be calculated using Pythagoras' theorem:

\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (10)^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5} \, \text{m/s} \]

### Step 6: Calculate the maximum height attained by the second part
Now, we need to find the maximum height attained by the second part after the explosion. The maximum height can be calculated using the formula:

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

Where \(v_y\) is the vertical component of the velocity after the explosion:

\[ h = \frac{(10)^2}{2 \cdot 10} = \frac{100}{20} = 5 \, \text{m} \]

### Final Result
The maximum height attained by the other part after the explosion is \(5 \, \text{m}\).

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