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A particle is projected from the ground so as to graze the four upper vertices of a regular hexagor, whose side is 2a and which is placed vertically with one side on the ground. What is the range on the ground. |
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Answer» `3sqrt(7A)`  `x_(0)=(R(4a cos 60^(@)+2a))/(2)` `x_(0)=(R)/(2)-2a` `y=TAN thetax(1-(x)/(R))` `asqrt(3)=tantheta x[(1-((R)/(2)-2a))/(R)]` `2asqrt(3)=tan theta[((R)/(2)-2X)]+2a cos 60^(@)[1-((R)/(2)-2a+a)/(R)]` `(1)/(2)=((R-4a))/(R-2a)([R-(R)/(2)+2a])/([R-(R)/(2)a])=((R-4a))/(((P,-2a)))` `([R+4a])/((R+2a))` `(1)/(2)=(R^(2)-16a^(2))/(R^(2)-4a^(2))` `R^(2)-4a^(2)=4R^(2)-32a^(2)` `R=2a sqrt(7)` |
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