Question : 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 (where 𝑔 is acceleration due to gravity)
(A) \dfrac{4v^{2}}{5g}
(B) \dfrac{4g}{5v^{2}}
(C) \dfrac{v^{2}}{g}
(D) \dfrac{4v^{2}}{\sqrt{5}g}
Answer : option (A)
Solution :

