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The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomena such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing. In the previous problem, what should be the corresponding projection angle.A. `tan^(-1) (1//2)`B. `tn^(-1)(1//3)`C. `tan^(-1)2`D. `tan^(-1)3` |
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Answer» Correct Answer - C `H = R tan theta - (1)/(2) g. (R^(2))/(u^(2)) (1 + tan^(2) theta)` `dy // d theta = 0` `tan theta = (u^(2))/(Rg) = (100)/(5 xx 10) = 2 rArr theta = tan^(-1) 2` |
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