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A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)A. `(4v^(2))/(5g)`B. `(4g)/(5v^(2))`C. `(v^(2))/(g)`D. `(4v^(2))/(sqrt (5)g)` |
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Answer» Correct Answer - 1 `R=2H` given Let `theta`=angle of projection `(2v_(x)v_(y))/(g)=2.(v_(y)^(2))/(2g)` `(v_(y))/(v_(x))=2` `tantheta=2` From tringle we can say that `sintheta=(2)/(sqrt(5)),costheta=(1)/(sqrt(5))` `:.` Range of projectile `R=(2v^(2)sintheta costheta)/(g)` `=(2v^(2))/(g)xx(2)/(sqrt(5))xx(1)/(sqrt(5))=(4v^(2))/(5g)` |
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