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A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is |
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Answer» `(v^(2))/(2g)`  `R= 2H rArr (v^(2)sin 2 theta_(0))/(g)= (2v^(2) sin^(2) theta_(0))/(2g) rArr COS theta_(0)= (1)/(2)` `rArr cos theta_(0)= (1)/(SQRT5)` `sin theta_(0)= (2)/(sqrt5) therefore R= (2v^(2) sin theta_(0)cos theta_(0))/(g)= (2v^(2) xx (1)/(sqrt5) (2)/(sqrt5))/(g)= (4v^(2))/(5g)` |
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