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

R = 2H given
We know R = 4H cot $\theta$ $\Rightarrow$ cot $\theta$ = $\frac{1}{2}$
From triangle we can say that sin $\theta$ = $\frac{2}{\sqrt{5}}$, cos $\theta$ = $\frac{1}{\sqrt{5}}$
$\therefore$ Range of projectile R = $\frac{2v^{2}sin\theta cos\theta }{g}$
= $\frac{2v^2}{g}$ $\times$ $\frac{2}{\sqrt{5}}$ $\times$ $\frac{1}{\sqrt{5}}$ = $\frac{4v^2}{5g}$