AP EAMCET · PHYSICS · Motion In Two Dimensions
The path of a projectile is given by the equation \(y=a x-b x^2\), where \(a\) and \(b\) are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively
- A \(\frac{2 a^2}{b}, \tan ^{-1}(a)\)
- B \(\frac{b^2}{2 a}, \tan ^{-1}(b)\)
- C \(\frac{a^2}{b}, \tan ^{-1}(2 b)\)
- D \(\frac{a^2}{4 b}, \tan ^{-1}(a)\)
Answer & Solution
Correct Answer
(D) \(\frac{a^2}{4 b}, \tan ^{-1}(a)\)
Step-by-step Solution
Detailed explanation
The given equation, \[ y=a x-b x^2 \] and we know that equation of trajectory is \[ y=(\tan \theta) x-\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta} \cdot x^2 \] Compare both equations, we get \[ a=\tan \theta, b=\frac{1}{2} \cdot \frac{g}{u^2 \cos ^2 \theta} \] The maximum height…
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