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 is acceleration due to gravity)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

R = 2 H given

$WeknowR=4Hcot\theta \Rightarrow cot\theta =\frac{1}{2}$
From triangle we can say that $sin\theta =\frac{2}{\sqrt{5}},cos\theta =\frac{1}{\sqrt{5}}\therefore RangeofprojectileR=\frac{2v^{2}sin\theta cos\theta }{g}$

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