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EEJ MAIN Physics QUESTION #7095
Question 1
A projectile trajectory is $y = \alpha x - \beta x^2$. Find the angle of projection and max height $H$.
  • $tan^{-1}\alpha$, $\frac{4\alpha^{2}}{\beta}$
  • $tan^{-1}\beta$, $\frac{\alpha^{2}}{2\beta}$
  • $tan^{-1}(\frac{\beta}{\alpha})$, $\frac{\alpha^{2}}{\beta}$
  • $tan^{-1}\alpha$, $\frac{\alpha^{2}}{4\beta}$✔️
Correct Answer Explanation

Compare $y = \alpha x - \beta x^2$ with $y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}$.

  • $\tan\theta = \alpha \implies \theta = \tan^{-1}\alpha$.
  • Max height occurs at $x = \frac{\alpha}{2\beta}$ (where $dy/dx = 0$).
  • $H = \alpha(\frac{\alpha}{2\beta}) - \beta(\frac{\alpha}{2\beta})^2 = \frac{\alpha^2}{2\beta} - \frac{\alpha^2}{4\beta} = \frac{\alpha^2}{4\beta}$.