JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 :-
- A \(\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}\)
- B \(\tan ^{-1} \beta, \frac{\alpha^{2}}{2 \beta}\)
- C \(\tan ^{-1} \alpha, \frac{4 \alpha^{2}}{\beta}\)
- D \(\tan ^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^{2}}{\beta}\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}\)
Step-by-step Solution
Detailed explanation
\(y =\alpha x -\beta x ^{2}\) comparing with trajectory equation \(y = x \tan \theta-\frac{1}{2} \frac{ gx ^{2}}{ u ^{2} \cos ^{2} \theta}\) \(\tan \theta=\alpha \Rightarrow \theta=\tan ^{-1} \alpha\) \(\beta=\frac{1}{2} \frac{ g }{ u ^{2} \cos ^{2} \theta}\)…
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