Kinematics Question 624
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 } $
Show Answer
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
