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At what angle `theta_(0)` with the horizontal, should a shell be fired if at the top of its trajectory its path has a radius of curvature equal to twice the maximum height of the trajectory |
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Answer» Correct Answer - `0045` At maximum height `R=(u^(2)cos^(2)theta_(0))/g` Maximum height of projectile `H=(u^(2)sin^(2)theta_(0))/(2g)` Given `R= 2H` `(u^(2) cos^(2)theta)/g=2((u^(2)sin^(2)theta)/(2g))Rightarrow tantheta=1 Rightarrow theta=45^(@)` |
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