MCQOPTIONS
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| 1. |
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 |
| A. | \[\frac{{{u}^{2}}{{\cos }^{2}}\theta }{g}\] |
| B. | \[\frac{{{u}^{2}}{{\cot }^{2}}\theta }{g\sin \theta }\] |
| C. | \[\frac{{{u}^{2}}}{g}\] |
| D. | \[\frac{{{u}^{2}}{{\tan }^{2}}\theta }{g\cos \theta }\] |
| Answer» C. \[\frac{{{u}^{2}}}{g}\] | |