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Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane
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The trajectory of a projectile in a vertical plane is $y=ax-bx^2$ where $a$ and $b$ are constants. The maximum height attained by the particle and the angle of projection from horizontal are

\[(a)\;\frac{b^2}{2a}, \tan ^{-1} (b)\quad (b)\;\frac{a^2}{4b},\tan ^{-1}(a) \quad (c)\;\frac{a^2}{b},\tan ^{-1} (2a) \quad (d)\;\frac{2a^2}{b},\tan ^{-1}(a)\]

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