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# What is the Radius of Curvature of the Parabola Traced Out by the Projectile in the Previous Problem at a Point Where the Particle Velocity Makes an Angle θ/2 with the Horizontal? - Physics

ConceptCircular Motion

#### Question

What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?

#### Solution

Let u be the initial velocity and v be the velocity at the point where it makes an angle $\frac{\theta}{2}$ with the horizontal component. It is given that the horizontal component remains unchanged.
Therefore, we get :

$v \cos \left( \frac{\theta}{2} \right) = u cos\theta$

$\Rightarrow v = \frac{u\cos\theta}{\cos\frac{\theta}{2}} . . . \left( i \right)$

$mg\cos\frac{\theta}{2} = \frac{m v^2}{r} . . . \left( ii \right)$

$\Rightarrow r = \frac{v^2}{g\cos\frac{\theta}{2}}$

On substituting the value of v from equation (i), we get :

$r = \frac{u^2 \cos^2 \theta}{g \cos^2 \frac{\theta}{2}}$

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Solution What is the Radius of Curvature of the Parabola Traced Out by the Projectile in the Previous Problem at a Point Where the Particle Velocity Makes an Angle θ/2 with the Horizontal? Concept: Circular Motion.
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