A particle is projected in an x-y plane with the y-axis along the vertical, the point of projection is the origin. The equation of a path is y = √3 x - (g x²)/2. Find the angle of projection and speed of projection.

To find the angle of projection and the speed of projection for the given path equation \( y = \sqrt{3}x - \frac{g x^2}{2} \), we can analyze the equation step by step.

Identifying the Components

The equation can be compared to the standard form of a projectile's trajectory, which is:

y = mx - (g/2v^2)x²

Here, \( m \) represents the slope of the trajectory, and \( g \) is the acceleration due to gravity.

Finding the Angle of Projection

The slope \( m \) is equal to \( \tan(\theta) \), where \( \theta \) is the angle of projection. From the equation:

• Given \( m = \sqrt{3} \)
• Thus, \( \tan(\theta) = \sqrt{3} \)

This implies:

• Angle \( \theta = 60^\circ \)

Calculating the Speed of Projection

Next, we need to find the speed of projection \( v \). From the equation, we can relate the coefficients:

• Comparing \( -\frac{g}{2v^2} \) with the coefficient of \( x^2 \), we have:
• \(-\frac{g}{2v^2} = -\frac{g}{2}\)

From this, we can deduce:

• Setting the two expressions equal gives us \( v^2 = 1 \)
• Thus, \( v = 1 \) (assuming \( g = 1 \) for simplicity)

Final Results

In summary:

• Angle of Projection: 60 degrees
• Speed of Projection: 1 unit (assuming \( g = 1 \))