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)
- A \(\frac{4 v^2}{5 g}\)
- B \(\frac{4 g}{5 v^2}\)
- C \(\frac{v^2}{g}\)
- D \(\frac{4 v^2}{\sqrt{5} g}\)
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,…
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