A Body Is Projected With Velocity U So That Its Horizontal Range Is Twice The Greatest Height Attained. The Value Of Range Is

The initial velocity of the projectile be u

Angle of projection = \( \Theta \)

As given in the question,

R = 2H

\( \Rightarrow \frac{u^{2}sin\Theta}{g} \) \( \Rightarrow 2 \frac{u^{2}sin{2}\Theta}{2g} \) \( 2 sin\Theta * cos \Theta = sin \Theta * sin \Theta \) \( tan \Theta = 2 \) \( sin \Theta = \frac{2}{\sqrt{5}} \) \( cos \Theta = \frac{1}{\sqrt{5}} \)

Range – R =\( \frac{2u^{2} sin\Theta * cos\Theta}{g} \)

R =\( \frac{2u^{2}}{g} \frac{2}{\sqrt{5}} * \frac{1}{\sqrt{5}} \)

R =\( \frac{4u^{2}}{5g} \)

Therefore, the value of range is \( \frac{4u^{2}}{5g} \)

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