Question : The trajectory of a projectile in vertical plane in 𝑦 = 𝑎𝑥 − 𝑏𝑥2 , where 𝑎 and 𝑏 are constant 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) \dfrac{b^{2}}{4b},\tan ^{-1}\left( b\right)
(B) \dfrac{a^{2}}{b},\tan ^{-1}\left( 2b\right)
(C) \dfrac{a^{2}}{4b},\tan ^{-1}\left( a\right)
(D) \dfrac{2a^{2}}{4b},\tan ^{-1}\left( a\right)
Answer : option (C)
Solution :
