A projectile is fired from the origin with an initial velocity v0 at an angle theta with the horizontal. The radius of curvature of its trajectory at the highest point is given by?

B. v_0^2 ^2 g — At the highest point, the velocity is purely horizontal, v = v0 cos(theta). The acceleration is g, acting perpendicular to the velocity, so R = v^2 / a_normal.

JEE Advanced · PhysicsHard

$A projectile is fired from the origin with an initial velocity v_{0} at an angle θ with the horizontal. The radius of curvature of its trajectory at the highest point is given by?$

A. $\frac{v _{0}^{2} s i n ^{2} θ}{g}$
B. $\frac{v _{0}^{2} c o s ^{2} θ}{g}$
C. $\frac{v _{0}^{2} t a n ^{2} θ}{g}$
D. $\frac{v _{0}^{2} s i n θ}{g}$

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Correct answer: B — $\frac{v _{0}^{2} c o s ^{2} θ}{g}$

Solution

At the highest point, the velocity is v = v_{0} cos θ . The normal acceleration is g, so the radius of curvature is R = \frac{v ^{2}}{g} = \frac{v _{0}^{2} cos ^{2} θ}{g} .

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A projectile is fired from the origin with an initial velocity v0​ at an angle θ with the horizontal. The radius of curvature of its trajectory at the highest point is given by?

Solution

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$A projectile is fired from the origin with an initial velocity v_{0} at an angle θ with the horizontal. The radius of curvature of its trajectory at the highest point is given by?$