JEE Main · PhysicsHard
A projectile is fired from the origin with an initial velocity $v_0$ at an angle $\theta$ with the horizontal. The radius of curvature of its trajectory at the highest point is given by?
- A.$\frac{v_0^2 \sin^2 \theta}{g}$
- B.$\frac{v_0^2 \cos^2 \theta}{g}$
- C.$\frac{v_0^2 \tan^2 \theta}{g}$
- D.$\frac{v_0^2 \sin \theta}{g}$
Show correct answer & step-by-step solution
Correct answer: B — $\frac{v_0^2 \cos^2 \theta}{g}$
Solution
At the highest point, the velocity is . The normal acceleration is , so the radius of curvature is .
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