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3-2.Motion in Plane
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પ્રક્ષિપ્ત પદાર્થની અવધિ ઊંચાઇ કરતાં બમણી હોય,તો અવધિ કેટલી હશે?
A
$\frac{{4{v^2}}}{{5g}}$
B
$\frac{{4g}}{{5{v^2}}}$
C
$\frac{{{v^2}}}{g}$
D
$\frac{{{4v^2}}}{\sqrt{5g}}$
Solution
$R = 4H\cot \theta $ $2H = 4H\cot \theta $
$\Rightarrow \,\cot \theta = \frac{1}{2}\,$; $\sin \theta = \frac{2}{{\sqrt 5 }}$; $\cos \theta = \frac{1}{{\sqrt 5 }}$
${\rm{Range}} = \frac{{{u^2}\,.\,2.\,\sin \theta \,.\,\cos \theta }}{g}$
$\therefore R = \frac{{2{u^2}\frac{2}{{\sqrt 5 }}\,.\,\frac{1}{{\sqrt 5 }}}}{g}$
$\therefore R = \frac{{4{u^2}}}{{5g}}$
Standard 11
Physics
