If R and H are horizontal range and maximum height attained by a projectile, than its speed of proje

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3-2.Motion in Plane

hard

If $R$ and $H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is ..........

$\sqrt{2 g R+\frac{4 R^2}{g H}}$

$\sqrt{2 g H+\frac{R^2 g}{8 H}}$

$\sqrt{2 g H+\frac{8 H}{R g}}$

$\sqrt{2 g H+\frac{R^2}{H}}$

(b)

$H=\frac{u^2 \sin ^2 \theta}{2 g} \Rightarrow \sin \theta=\sqrt{\frac{2 g H}{u^2}}$

$R=\frac{2 u^2 \sin \theta \cos \theta}{g}$

$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} \times \sqrt{1-\frac{2 g H}{u^2}}$

$R=\frac{2 u^2}{g} \sqrt{\frac{2 g H}{u^2}} \times \sqrt{\frac{u^2-2 g H}{u^2}}$

$\frac{g R}{2 \sqrt{2 g H}}=\sqrt{u^2-2 g H}$

Squaring both the sides,

$\frac{g R^2}{4 \times 2 g H}=u^2-2 g H$

$\Rightarrow u^2=2 g H+\frac{9 R^2}{8 H}$

$u=\sqrt{2 g H+\frac{g R^2}{8 H}}$

Standard 11

Physics

hard

(AIPMT-2011)

easy

medium

medium

medium