AP EAMCET · PHYSICS · Motion In Two Dimensions
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}
- A A-i,B-iii, C-iv, D-ii
- B A-i,B-iii, C-ii, D-iv
- C A-iii,B-i, C-iv, D-ii
- D A-iv,B-ii, C-iii, D-i
Answer & Solution
Correct Answer
(A) A-i,B-iii, C-iv, D-ii
Step-by-step Solution
Detailed explanation
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…
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