Kinematics Question 619

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

Options:

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

B)$ \frac{4g}{5v^{2}} $

C) $ \frac{v^{2}}{g} $

D)$ \frac{4v^{2}}{\sqrt{5}g} $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] We know, $ R=4H\cot \theta \Rightarrow \cot \theta =\frac{1}{2} $

$ \text{From triangle we can say that } $

$ \sin \theta =\frac{2}{\sqrt{5}},\text{ cos}\theta \text{=}\frac{1}{\sqrt{5}} $

$ \therefore $ Range of projectile $ R=\frac{2v^{2}\sin \theta \cos \theta }{g} $

$ =\frac{2v^{2}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4v^{2}}{5g} $



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