Path of projectile is given by the equation $Y=P x-Q x^2$, match the following accordingly (acceleration due…

Path of projectile is given by the equation $Y=P x-Q x^2$, match the following accordingly (acceleration due to gravity = g) $\begin{array}{|l|l|l|l|}\hline A & \text{Range} & (i) & \frac{P}{Q} \\\hline B & \text{Maximum height} & (ii) & P \\\hline C & \text{Time of flight} & (iii) & \frac{P^{2}}{4 Q} \\\hline D & \text{Tangent of projection} & (iv) & \left(\sqrt{\frac{2}{g Q}}\right) P \\\hline\end{array}$
  1. A-i,B-iii, C-iv, D-ii
  2. A-i,B-iii, C-ii, D-iv
  3. A-iii,B-i, C-iv, D-ii
  4. A-iv,B-ii, C-iii, D-i

Solution

Path of projectile, $\mathrm{Y}=\mathrm{Px}-\mathrm{Q} \mathrm{x}^2$ $\begin{array}{ll} \therefore & \text { At } x=R, Y=0 \\ \Rightarrow & 0=P \cdot R-Q R^2 \Rightarrow R(P-Q R)=0 \end{array}$ $\therefore \quad$ Range, $\mathrm{R}=\frac{\mathrm{P}}{\mathrm{Q}}$ At $\mathrm{x}=\frac{\mathrm{R}}{2}, \mathrm{Y}=\mathrm{H}$ $\therefore \quad$ Maximum Height, $\mathrm{H}=\mathrm{P}\left(\frac{\mathrm{R}}{2}\right)-\mathrm{Q}\left(\frac{\mathrm{R}}{2}\right)^2$ $\begin{aligned} & =P\left(\frac{P}{2 Q}\right)-Q\left(\frac{P}{2 Q}\right)^2 \\ & =\frac{P^2}{2 Q}-\frac{P^2}{4 Q}=\frac{P^2}{4 Q} \end{aligned}$
Tangent of projection, $\tan \theta=\mathrm{P}$ Also, $\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{Q} \Rightarrow \mathrm{u} \cos \theta=\sqrt{\frac{\mathrm{g}}{2 \mathrm{Q}}}$ $\therefore \mathrm{R}=\mathrm{T}(\mathrm{ucos} \theta) \Rightarrow \mathrm{T}=\frac{\frac{\mathrm{P}}{\mathrm{Q}}}{\sqrt{\frac{\mathrm{~g}}{2 \mathrm{Q}}}}=\left(\sqrt{\frac{2}{\mathrm{gQ}}}\right) \mathrm{P}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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