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.
• B.
• C.
• D.

Answer 1

The correct option is A

We know $R=4Hcot\theta$
$2H=4Hcot\theta \Rightarrow cot\theta =\frac{1}{2};sin\theta =\frac{2}{\sqrt{5}};cos\theta =\frac{1}{\sqrt{5}}[AsR=2Hgiven]Range=\frac{u^2.2.sin\theta .cos\theta }{g}=\frac{2u^2\frac{2}{\sqrt{5}}.\frac{1}{\sqrt{5}}}{g}=\frac{4u^2}{5g}$