ExamBro
ExamBro
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

  1. A \(\frac{Q^2}{2 P}\)
  2. B \(\frac{P^2}{Q}\)
  3. C \(4 P\)
  4. D \(\frac{P}{4}\)
Verified Solution

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…