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AP EAMCET · PHYSICS · Motion In Two Dimensions

A particle is projected with a velocity \(v\) such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is \((g=\) acceleration due to gravity)

  1. A \(\frac{4 v^2}{5 g}\)
  2. B \(\frac{4 g}{5 v^2}\)
  3. C \(\frac{v^2}{g}\)
  4. D \(\frac{4 v^2}{\sqrt{5} g}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{4 v^2}{5 g}\)

Step-by-step Solution

Detailed explanation

Velocity of particle \(=v\) If \(\theta\) is angle of projection such a way,…