Physics Two Dimensional Motion Question 172
Question: If a particle is projected with speed u from ground at an angle with horizontal, then radius of curvature of a point where velocity vector is perpendicular to initial velocity vector is given by
Options:
A) $ \frac{u^{2}{{\cos }^{2}}\theta }{g} $ B)$ \frac{u^{2}{{\cot }^{2}}\theta }{g\sin \theta } $ C) $ \frac{u^{2}}{g} $ D)$ \frac{u^{2}{{\tan }^{2}}\theta }{g\cos \theta } $
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Answer:
Correct Answer: B
Solution:
[b] Horizontal components of velocity at O and P are equal. $ \
Therefore \text{v}\cos ( 90{}^\circ -\theta)=\text{u}\cos \theta $
$ \text{or },\text{v sin}\theta =u\cos \theta $
$ \text{or },\text{v},\text{=},,\text{u},\text{cos}\theta $
$ \text{At },\text{P, }\frac{\text{v} _{T}^{2}}{R}=a _{c}\text{ ;} $
$ \frac{{{u^{2}}{{\cot }^{2}}\theta }{g\sin \theta }=R $
BETA
