For a given velocity, a projectile has the same range R for two angles of projection if $t_{1}$ and $t_{2}$ are the times of flight in the two cases then

t_{1} t_{2} \propto R^{2}

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t_{1} t_{2} \propto R

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t_{1} t_{2} \propto \frac{1}{R}

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t_{1} t_{2} \propto \frac{1}{R^{2}}

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Solution

The correct option is B $t_{1} t_{2} \propto R$
For same range angle of projection should be $θ$ and $90 - θ$
So, time of flights $t_{1} = \frac{2 u s i n θ}{g}$ and
$t_{2} = \frac{2 u s i n (90 - θ)}{g} = \frac{2 u c o s θ}{g}$
By multiplying = $t_{1} t_{2} = \frac{4 u^{2} s i n θ c o s θ}{g^{2}}$
$t_{1} t_{2} = \frac{2}{g} \frac{(u^{2} s i n 2 θ)}{g} = \frac{2 R}{g} \Rightarrow t_{1} t_{2} \propto R$

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For a given velocity, a projectile has 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 B t1t2∝RFor same range angle of projection should be θ and 90−θ So, time of flights t1=2u sin θg and t2=2u sin(90−θ)g=2u cos θg By multiplying =t1t2=4u2 sin θ cos θg2 t1t2=2g(u2 sin 2θ)g=2Rg⇒t1t2∝R

For a given velocity, a projectile has the same range R for two angles of projection if $t_{1}$ and $t_{2}$ are the times of flight in the two cases then