A projectile has the same range for two angles of projection. If times of flight in both cases are $t_{1}$ and $t_{2}$ then the range of the projectile is :

\frac{1}{2} g t_{1} t_{2}

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\frac{1}{4} g t_{1} t_{2}

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g t_{1} t_{2}

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\frac{1}{8} g t_{1} t_{2}

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Solution

The correct option is D $\frac{1}{2} g t_{1} t_{2}$
Horizontal Range is same for complementry angle of projection i.e., for $θ$ & $(90 - θ)$
time of flight $= \frac{2 u sin θ}{g}$

$t h e r e f o r e t_{1} = \frac{2 u sin θ}{g}$

$t_{2} = \frac{2 u sin (90 - θ)}{g} = \frac{2 u cos θ}{g}$

Now, $t_{1} . t_{2} = (\frac{2 u sin θ}{g}) (\frac{2 u cos θ}{g})$

$= \frac{2 u^{2}}{g . g} (2 sin θ cos θ)$

$= \frac{2}{g} (\frac{u^{2} sin 2 θ}{g})$

$t_{1} . t_{2} = \frac{2}{g} R \Rightarrow R = \frac{1}{2} g t_{1} t_{2}$

$∴ R = \frac{1}{2} g t_{1} t_{2}$

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A projectile has the same range for two angles of projection. If times of flight in both cases are t1 and t2 then the range of the projectile is :

The correct option is D 12gt1t2Horizontal Range is same for complementry angle of projection i.e., for θ & (90−θ)time of flight =2usinθgtherefore t1=2usinθgt2=2usin(90−θ)g=2ucosθgNow, t1.t2=(2usinθg)(2ucosθg)=2u2g.g(2sinθcosθ)=2g(u2sin2θg)t1.t2=2gR⇒R=12g t1t2∴R=12g t1t2

A projectile has the same range for two angles of projection. If times of flight in both cases are $t_{1}$ and $t_{2}$ then the range of the projectile is :