Question

A particle is projected with a velocity $v$, so that its range on a horizontal plane is twice the greatest height attained. If $g$ is acceleration due to gravity, then its range is:

• A. $\frac{4v^2}{5g}$
• B. $\frac{4g}{5v^2}$
• C. $\frac{4v^3}{5g^2}$
• D. $\frac{4v}{5g^2}$

Answer 1

Answer: (A)

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

Range should be twice the greatest height

$R=2H$

$\frac{v^{2}sin2\theta }{g}=2\frac{v^{2}si{n}^2\theta }{2g}$

$⟹sin2\theta =si{n}^2\theta$

$2sin\theta cos\theta =si{n}^2\theta$

$\therefore tan\theta =2$

$sin\theta =\frac{2}{\sqrt{5}}$

therefore range is $R=\frac{v^{2}sin2\theta }{g}=\frac{2v^{2}sin\theta cos\theta }{g}=\frac{2v^2\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}}{g}=\frac{4v^2}{5g}$