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3-2.Motion in Plane
hard
સમાન અવધિ $R$ ધરાવતા બે પ્રક્ષિપ્તકોણે પદાર્થને ફેંકતા ઉડ્ડયન સમય અનુક્રમે $t_1$ અને $t_2$ મળે છે.તો નીચેનામાથી શું સાચું છે?
A${t_1}{t_2} \propto \,{R^2}$
B${t_1}{t_2} \propto \,R$
C${t_1}{t_2} \propto \,\frac{1}{R}$
D${t_1}{t_2} \propto \,\frac{1}{{{R^2}}}$
(AIEEE-2004) (AIEEE-2005) (AIIMS-2006)
Solution
(b) For same range angle of projection should be $\theta$ and $90-\theta$
So, time of flights ${t_1} = \frac{{2u\sin \theta }}{g}$ and
${t_2} = \frac{{2u\sin (90 – \theta )}}{g} = \frac{{2u\cos \theta }}{g}$
By multiplying$ = {t_1}{t_2} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}}$
${t_1}{t_2} = \frac{2}{g}\frac{{({u^2}\sin 2\theta )}}{g} = \frac{{2R}}{g}$
$⇒$ ${t_1}{t_2} \propto R$
So, time of flights ${t_1} = \frac{{2u\sin \theta }}{g}$ and
${t_2} = \frac{{2u\sin (90 – \theta )}}{g} = \frac{{2u\cos \theta }}{g}$
By multiplying$ = {t_1}{t_2} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}}$
${t_1}{t_2} = \frac{2}{g}\frac{{({u^2}\sin 2\theta )}}{g} = \frac{{2R}}{g}$
$⇒$ ${t_1}{t_2} \propto R$
Standard 11
Physics
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