Question:
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 :
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 :
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Compare with \(y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}\); the slope at the origin is \(\tan\theta\), and H is the maximum of y.
Updated On: Nov 4, 2026
- \(\tan^{-1}\alpha,\ \frac{\alpha^2}{4\beta}\)
- \(\tan^{-1}\alpha,\ \frac{4\alpha^2}{\beta}\)
- \(\tan^{-1}\left(\frac{\beta}{\alpha}\right),\ \frac{\alpha^2}{\beta}\)
- \(\tan^{-1}\beta,\ \frac{\alpha^2}{2\beta}\)
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The Correct Option is A
Solution and Explanation
The standard trajectory is \(y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}x^2\). Comparing with \(y = \alpha x - \beta x^2\): \(\tan\theta = \alpha\), so \(\theta = \tan^{-1}\alpha\), and \(\beta = \frac{g}{2u^2\cos^2\theta}\).
\(H = \frac{u^2\sin^2\theta}{2g} = \tan^2\theta \cdot \frac{u^2\cos^2\theta}{2g} = \alpha^2 \cdot \frac{1}{4\beta} = \frac{\alpha^2}{4\beta}\).
Answer: (1) \(\tan^{-1}\alpha,\ \frac{\alpha^2}{4\beta}\)
\(H = \frac{u^2\sin^2\theta}{2g} = \tan^2\theta \cdot \frac{u^2\cos^2\theta}{2g} = \alpha^2 \cdot \frac{1}{4\beta} = \frac{\alpha^2}{4\beta}\).
Answer: (1) \(\tan^{-1}\alpha,\ \frac{\alpha^2}{4\beta}\)
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