Q123 / 160 · AP EAMCET 2024 (23 May Shift 1) · Physics
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 & Range & (i) & \frac{P}{Q} \\\hline B & Maximum height & (ii) & P \\\hline C & Time of flight & (iii) & \frac{P^2}{4 Q} \\\hline D & Tangent of projection & (iv) & \left(\sqrt{\frac{2}{g Q}}\right) P \\\hline\end{array}
- Correct answerAA-i,B-iii, C-iv, D-ii
- BA-i,B-iii, C-ii, D-iv
- CA-iii,B-i, C-iv, D-ii
- DA-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, $\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{Q} \Rightarrow \mathrm{u} \cos \theta=\sqrt{\frac{\mathrm{g}}{2 \mathrm{Q}}}$ $\therefore \mathrm{R}=\mathrm{T}(\mathrm{ucos} \theta) \Rightarrow \mathrm{T}=\frac{\frac{\mathrm{P}}{\mathrm{Q}}}{\sqrt{\frac{\mathrm{~g}}{2 \mathrm{Q}}}}=\left(\sqrt{\frac{2}{\mathrm{gQ}}}\right) \mathrm{P}$
Tangent of projection, $\tan \theta=\mathrm{P}$ Also, $\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\mathrm{Q} \Rightarrow \mathrm{u} \cos \theta=\sqrt{\frac{\mathrm{g}}{2 \mathrm{Q}}}$ $\therefore \mathrm{R}=\mathrm{T}(\mathrm{ucos} \theta) \Rightarrow \mathrm{T}=\frac{\frac{\mathrm{P}}{\mathrm{Q}}}{\sqrt{\frac{\mathrm{~g}}{2 \mathrm{Q}}}}=\left(\sqrt{\frac{2}{\mathrm{gQ}}}\right) \mathrm{P}$
Hence, the correct answer is option: 'A'.