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A particle is projected with speed `u` at angle `theta` to the horizontal. Find the radius of curvature at highest point of its trajectoryA. `(u^(@)cos^(2)theta)/(2g)`B. `(sqrt(3)u^(2)cos^(2)theta)/(2g)`C. `(u^(2)cos^(2)theta)/(g)`D. `(sqrt(3)u^(2)cos^(2)theta)/(g)` |
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Answer» Correct Answer - c `r=(u^(2)cos^(2)theta)/(g)` |
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