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3-2.Motion in Plane
hard
જો $R$ અને $H$ એ સમક્ષિતિજ અવધિ અને મહત્તમ ઊંચાઈ હોય તો તેના પ્રક્ષેપણની ઝડપ શોધો
A
$\sqrt{2 g R+\frac{4 R^2}{g H}}$
B
$\sqrt{2 g H+\frac{R^2 g}{8 H}}$
C
$\sqrt{2 g H+\frac{8 H}{R g}}$
D
$\sqrt{2 g H+\frac{R^2}{H}}$
Solution
(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
