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A stone is thrown horizontally with a velocity of `10m//sec`. Find the radius of curvature of it’s trajectory at the end of `3 s` after motion began. `(g=10m//s^(2))`A. `10sqrt10m`B. `100sqrt10m`C. `sqrt10m`D. `100 m` |
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Answer» Correct Answer - A Radius of curvature `r=v_(t)^(2)/(g sin alpha)` `V=sqrt(v_(x)^(2)+v_(y)^(2))=10sqrt10` `tan alpha=v_(x)/v_(y)=1/3, sin alpha=1/sqrt10` |
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