A projectile can have the same range R for two angles of projection. If $t_{1} a n d t_{2}$ are the times of flight in the
two cases, then

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Solution

The correct option is A
$t_{1} = \frac{2 u s i n θ}{g}, t_{2} = \frac{2 u s i n (90 - θ)}{g} = \frac{2 u c o s θ}{g} t_{1} t_{2} = \frac{4 u^{2} s i n θ c o s θ}{g^{2}} = \frac{2}{g} (\frac{u^{2} s i n 2 θ}{g}) = \frac{2}{g} R H e n c e, t_{1} t_{2} \propto R$

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A projectile can have the same range R for two angles of projection. If t1 and t2 are the times of flight in the two cases, then

The correct option is A t1=2u sinθg,t2=2u sin(90−θ)g=2u cosθgt1t2=4u2sinθcosθg2=2g(u2sin 2θg)=2gRHence,t1t2∝ R

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The correct option is A
$t_{1} = \frac{2 u s i n θ}{g}, t_{2} = \frac{2 u s i n (90 - θ)}{g} = \frac{2 u c o s θ}{g} t_{1} t_{2} = \frac{4 u^{2} s i n θ c o s θ}{g^{2}} = \frac{2}{g} (\frac{u^{2} s i n 2 θ}{g}) = \frac{2}{g} R H e n c e, t_{1} t_{2} \propto R$