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  • The trajectory of a projectile in a vertical plane is , where <img alt="\alpha and \beta" src="https://entrance

The trajectory of a projectile in a vertical plane is y= \alpha x - \beta x^2, where \alpha and \beta are constants and x and y are repesctively 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:
Option: 1 \tan^{-1} \beta , \frac{\alpha ^{2} }{2 \beta}
Option: 2 \tan ^{-1}\alpha , \frac{4 \alpha ^2}{\beta }
Option: 3 \tan ^{-1}\alpha , \frac{\alpha ^2}{4 \beta }
Option: 4 \tan ^{-1} \left (\frac{\beta }{\alpha} \right ), \frac{ \alpha ^2}{\beta }

Answers (1)

best_answer

\mathrm{y}=\alpha \mathrm{x}-\beta \mathrm{x}^{2}

comparing with trajectory equation
\begin{array}{l} y=x \tan \theta-\frac{1}{2} \frac{g x^{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} \end{array}

Maximum height : H
\begin{array}{l} \mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{g}}{2 \beta \cos ^{2} \theta} \frac{\sin ^{2} \theta}{2 \mathrm{~g}} \\ \\ \mathrm{H}=\frac{\tan ^{2} \theta}{4 \beta}=\frac{\alpha^{2}}{4 \beta} \end{array}
 

Posted by

Deependra Verma

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