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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 :-

  1. A \(\tan ^{-1} \alpha, \frac{\alpha^{2}}{4 \beta}\)
  2. B \(\tan ^{-1} \beta, \frac{\alpha^{2}}{2 \beta}\)
  3. C \(\tan ^{-1} \alpha, \frac{4 \alpha^{2}}{\beta}\)
  4. D \(\tan ^{-1}\left(\frac{\beta}{\alpha}\right), \frac{\alpha^{2}}{\beta}\)
Verified Solution

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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