Two particles are simultaneously projected in the same vertical plane from the same point with velocities u and v at angles `alpha and beta` with horizontal. Find the time that elapses when their velocities are parallel.

Explanation

To find the time that elapses when the velocities of two particles projected at angles \(\alpha\) and \(\beta\) become parallel, we can follow these steps:

### Step 1: Write the velocity equations for both particles
The velocity of the first particle \( \vec{v_1} \) at time \( t \) can be expressed as:
\[ \vec{v_1} = u \cos \alpha \hat{i} + (u \sin \alpha - gt) \hat{j} \]
where \( u \) is the initial velocity of the first particle, \( \alpha \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Similarly, the velocity of the second particle \( \vec{v_2} \) at time \( t \) is:
\[ \vec{v_2} = v \cos \beta \hat{i} + (v \sin \beta - gt) \hat{j} \]
where \( v \) is the initial velocity of the second particle and \( \beta \) is the angle of projection.

### Step 2: Set the condition for parallel velocities
For the velocities to be parallel, the direction ratios of the velocities must be equal. This gives us the following relationship:
\[ \frac{u \cos \alpha}{v \cos \beta} = \frac{u \sin \alpha - gt}{v \sin \beta - gt} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying the equation gives:
\[ u \cos \alpha (v \sin \beta - gt) = v \cos \beta (u \sin \alpha - gt) \]

### Step 4: Expand and rearrange the equation
Expanding both sides results in:
\[ u v \cos \alpha \sin \beta - u \cos \alpha gt = v u \sin \alpha \cos \beta - v \cos \beta gt \]
Rearranging this gives:
\[ u v \cos \alpha \sin \beta - v u \sin \alpha \cos \beta = gt (u \cos \alpha + v \cos \beta) \]

### Step 5: Factor out common terms
The left side can be factored as:
\[ uv (\cos \alpha \sin \beta - \sin \alpha \cos \beta) = gt (u \cos \alpha + v \cos \beta) \]

### Step 6: Use the sine difference identity
Using the sine difference identity, we can simplify this to:
\[ uv \sin(\alpha - \beta) = gt (u \cos \alpha + v \cos \beta) \]

### Step 7: Solve for time \( t \)
Now, we can solve for \( t \):
\[ t = \frac{uv \sin(\alpha - \beta)}{g (u \cos \alpha + v \cos \beta)} \]

### Final Answer
Thus, the time that elapses when the velocities of the two particles are parallel is:
\[ t = \frac{uv \sin(\alpha - \beta)}{g (u \cos \alpha + v \cos \beta)} \]

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