AP EAMCET · PHYSICS · Motion In Two Dimensions
Equation of a projectile is given by \(y=P x-Q x^2\), where \(P\) and \(Q\) are constants. The ratio of maximum height to the range of the projectile is
- A \(\frac{Q^2}{2 P}\)
- B \(\frac{P^2}{Q}\)
- C \(4 P\)
- D \(\frac{P}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{P}{4}\)
Step-by-step Solution
Detailed explanation
We have, \(y=P x-Q x^2\) ...(i) Equation of trajectory is \(y=x \tan \theta-\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \theta}\) ...(ii) After comparing Eqs. ( \(i)\) and (ii), we get \(P=\tan \theta\) ...(iii) \(Q=\frac{g}{2 u^2 \cos ^2 \theta}\) ...(iv) We know that…
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