The trajectory of a projectile in a vertical plane is \(y =\alpha x -\beta x ^{2},\) where \(\alpha\) and \(\beta\) are constants and \(x \& y\) are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection \(\theta\) and the maximum height attained \(H\) are respectively given by :-

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(A) \(\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}\)

Detailed explanation

\(y =\alpha x -\beta x ^{2}\)

comparing with trajectory equation

\(y = x \tan \theta-\frac{1}{2} \frac{ gx ^{2}}{ u ^{2} \cos ^{2} \theta}\)

\(\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1} \alpha\)

\(\beta=\frac{1}{2} \frac{ g }{ u ^{2} \cos ^{2} \theta}\)

\(u ^{2}=\frac{ g }{2 \beta \cos ^{2} \theta}\)

Maximum height : \(H\)

\(H =\frac{ u ^{2} \sin ^{2} \theta}{2 g }=\frac{ g }{2 \beta \cos ^{2} \theta} \frac{\sin ^{2} \theta}{2 g }\)

\(H =\frac{\tan ^{2} \theta}{4 \beta}=\frac{\alpha^{2}}{4 \beta}\)

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