a particle is projected with velocity u at an anglep with the horizontal. find the radius of the curvature traced out by the parabola at the point where veocity makes an angle p/2 with the horizontal

To find the radius of curvature of the trajectory traced by a particle projected at an angle with the horizontal, we need to analyze the motion in two dimensions: horizontal and vertical. The trajectory of the particle is a parabola, and we can derive the radius of curvature at the point where the velocity vector makes an angle of \( \frac{\pi}{2} \) (or 90 degrees) with the horizontal.

Understanding the Motion

When a particle is projected with an initial velocity \( u \) at an angle \( \theta \) to the horizontal, its horizontal and vertical components of velocity can be expressed as:

• Horizontal component: \( u_x = u \cos(\theta) \)
• Vertical component: \( u_y = u \sin(\theta) - gt \) (where \( g \) is the acceleration due to gravity)

Velocity at the Desired Angle

We need to find the point in the trajectory where the velocity vector makes an angle of \( \frac{\pi}{2} \) with the horizontal. This occurs when the horizontal component of the velocity becomes zero, meaning:

At this point, \( u_x = 0 \). Since \( u_x = u \cos(\theta) \), this implies that the particle has reached its maximum height, where the vertical component of the velocity is at its peak.

Finding the Time of Flight to Maximum Height

The time \( t \) to reach maximum height can be calculated using the vertical motion equation:

At maximum height, the vertical component of velocity becomes zero:

0 = \( u \sin(\theta) - gt \)

From this, we can solve for \( t \):

t = \( \frac{u \sin(\theta)}{g} \)

Position at Maximum Height

Next, we need to find the position of the particle at this time. The horizontal and vertical displacements can be calculated as follows:

• Horizontal displacement: \( x = u \cos(\theta) \cdot t = u \cos(\theta) \cdot \frac{u \sin(\theta)}{g} = \frac{u^2 \sin(\theta) \cos(\theta)}{g} \)
• Vertical displacement: \( y = u \sin(\theta) \cdot t - \frac{1}{2} g t^2 = u \sin(\theta) \cdot \frac{u \sin(\theta)}{g} - \frac{1}{2} g \left( \frac{u \sin(\theta)}{g} \right)^2 \)

After simplifying, we find:

y = \( \frac{u^2 \sin^2(\theta)}{2g} \)

Radius of Curvature Formula

The radius of curvature \( R \) at any point in a trajectory can be calculated using the formula:

R = \( \frac{(1 + (dy/dx)^2)^{3/2}}{|d^2y/dx^2|} \)

To find \( dy/dx \) and \( d^2y/dx^2 \), we can use parametric equations for the trajectory:

• From the equations of motion, we can express \( y \) in terms of \( x \) and differentiate accordingly.

Final Calculation

At the maximum height, the slope \( dy/dx \) is zero, leading to a simplified radius of curvature. The radius of curvature at this point can be derived as:

R = \( \frac{u^2}{g} \)

This indicates that the radius of curvature at the point where the velocity is vertical is directly proportional to the square of the initial velocity and inversely proportional to the acceleration due to gravity.

In summary, the radius of curvature traced out by the parabola at the point where the velocity makes an angle of \( \frac{\pi}{2} \) with the horizontal is given by:

R = \( \frac{u^2}{g} \)