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A particle is projected with `10ms^(-1)` at angle `theta` above horizontal from a horizontal ground. Find the value of `theta` so that area under the parabolic path of the projectile will be maximum `(g=10ms^(-2))`.A. `60^(@)`B. `45^(@)`C. `90^(@)`D. None of these |
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Answer» `A=int y dx=int_(0)^(R)tan theta(x-(x^(2))/(R))dx` `=tan theta((R^(2))/(2)-(R^(3))/(3R))` `(dA)/(dtheta)=0=2 sin^(2)thetacos^(2)theta-sin^(3)thetasintheta=0` `=tan theta(R^(2))/(6)=(tantheta)/(6)((u^(2)sin theta cos theta)/(g))^(2)` `sin^(2)theta(3 cos^(2)theta-sin^(2)theta)=0` `=(10^(4)xxu^(2))/(36xx10xx10cos theta)sin^(2)thetacos^(2)theta` `3=tan^(2)theta` `=(200)/(3)sin^(3)theta cos theta` `tan theta=sqrt(3)` `theta=60^(@)` |
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