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Work Done as Dot Product

The trajector...

Question

The trajectory of a projectile in a vertical plane is $y = a x - b x^{2}$ , where $a$ and $b$ are constants and $x$ and $y$ are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

A

\frac{b^{2}}{2 a}, t a n^{- 1} (b)

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B

\frac{a^{2}}{b}, t a n^{- 1} (2 a)

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C

\frac{a^{2}}{4 b}, t a n^{- 1} (a)

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D

\frac{2 a^{2}}{b}, t a n^{- 1} (a)

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Solution

The correct option is C $\frac{a^{2}}{4 b}, t a n^{- 1} (a)$

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