A projectile has same range R for two angles of projection. If $t_{1} & t_{2}$ be the same time of flight for the two cases then:

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

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R = \frac{g {(t_{1} + t_{2})}_{1}^{2}}{2}

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R = g \sqrt{t_{1} t_{2}}

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R = 2 g \frac{t_{1} t_{2}}{t_{1} + t_{2}}

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Solution

The correct option is A $R = \frac{g t_{1} t_{2}}{2}$
$\begin{matrix} T h e s tan d a r d f o r m u l a f o r R a n g e o f p r o j e c t i l e i s R = \frac{u^{2} sin (2 α)}{g} . . . . . . . . . . . . . . . . . (1) T_{1} = \frac{2 u sin (α)}{g} T_{2} = \frac{2 u sin (90^{0} - α)}{g} T_{2} = \frac{2 u cos (α)}{g} T a k i n g p r o d u c t o f t w o t i m e s o f f l i g h t T_{1} T_{2} = \frac{2 u^{2} (2 sin (α)) cos (α)}{g^{2}} T_{1} T_{2} = \frac{2}{g} \frac{u^{2} sin (2 α)}{g} . . . . . . . . . . . . . (2) D i v i d i n g e q (1) b y (2) \frac{R}{T_{1} T_{2}} = \frac{g}{2} R = \frac{T_{1} T_{2} g}{2} \end{matrix}$

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A projectile has same range R for two angles of projection. If t1&t2 be the same time of flight for the two cases then:

The correct option is A R=gt1t22ThestandardformulaforRangeofprojectileisR=u2sin(2α)g.................(1)T1=2usin(α)gT2=2usin(900−α)gT2=2ucos(α)gTakingproductoftwotimesofflightT1T2=2u2(2sin(α))cos(α)g2T1T2=2gu2sin(2α)g.............(2)Dividingeq(1)by(2)RT1T2=g2R=T1T2g2

A projectile has same range R for two angles of projection. If $t_{1} & t_{2}$ be the same time of flight for the two cases then: