Question

A particle is projected from ground at an angle θ with horizontal with speed u. The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is?
Options are
A)1/sin^2θ cosθ B)cos^2θ
C)1/sin^3θ D)1/cos^3θ

Answer 1

Answer on Question #49916-Physics-Mechanics-Kinematics-Dynamics

A particle is projected from ground at an angle $\theta$ with horizontal with speed $u$ . The ratio of radius of curvature of its trajectory at point of projection to radius of curvature at maximum height is? Options are

A)1/sin^2θ cosθ B)cos^2θ C)1/sin^3θ D)1/cos^3θ

Solution

$$x (t) = u \cos \theta t, y (t) = u \sin \theta t - \frac {1}{2} g t ^ {2}.$$

Radius of curvature is

$$R = \left| \frac {\left[ \left(x ^ {\prime} (t)\right) ^ {2} + \left(y ^ {\prime} (t)\right) ^ {2} \right] ^ {3 / 2}}{x ^ {\prime} (t) y ^ {\prime \prime} (t) - y ^ {\prime} (t) x ^ {\prime \prime} (t)} \right|.$$ $$x ^ {\prime} (t) = u \cos \theta , x ^ {\prime \prime} (t) = 0.$$ $$y ^ {\prime} (t) = u \sin \theta - g t, y ^ {\prime \prime} (t) = - g.$$

At $t = 0$

$$x ^ {\prime} (0) = u \cos \theta , x ^ {\prime \prime} (0) = 0, y ^ {\prime} (0) = u \sin \theta , y ^ {\prime \prime} (0) = - g.$$ $$\mathrm {A t} t = \frac {u \sin \theta}{g}$$ $$x ^ {\prime} = u \cos \theta , x ^ {\prime \prime} = 0, y ^ {\prime} = 0, y ^ {\prime \prime} = - g.$$

Substituting values we get

$$R (t = 0) = \frac {u ^ {2}}{g \cos \theta}.$$ $$R \left(t = \frac {u \sin \theta}{g}\right) = \frac {u ^ {2} \cos^ {2} \theta}{g}.$$

The ratio is

$$\frac {R (t = 0)}{R \left(t = \frac {u \sin \theta}{g}\right)} = \frac {1}{\cos^ {3} \theta}.$$

Answer: D) $\frac{1}{\cos^3\theta}$

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