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(D) The horizontal range $R$ is given by $R = \frac{V^2 \sin(2	heta)}{g} = \frac{2V^2 \sin	heta \cos	heta}{g}$.The time taken to reach the highest

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$\cot^{-1}\left(\frac{R}{20t^2}\right)$

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(D) The horizontal range $R$ is given by $R = \frac{V^2 \sin(2	heta)}{g} = \frac{2V^2 \sin	heta \cos	heta}{g}$.The time taken to reach the highest point (where inclination is zero) is $t = \frac{V \sin	heta}{g}$.From this,$V \sin	heta = gt$,so $V = \frac{gt}{\sin	heta}$.Substitute $V$ into the range formula: $R = \frac{2(gt/\sin	heta)^2 \sin	heta \cos	heta}{g} = \frac{2g^2 t^2 \sin	heta \cos	heta}{g \sin^2	heta} = \frac{2gt^2 \cos	heta}{\sin	heta} = \frac{2gt^2}{	an	heta}$.Given $g = 10 \, m/s^2$,we have $R = \frac{2(10)t^2}{	an	heta} = \frac{20t^2}{	an	heta}$.Rearranging for $	heta$,we get $	an	heta = \frac{20t^2}{R}$,which implies $\cot	heta = \frac{R}{20t^2}$.Therefore,$	heta = \cot^{-1}\left(\frac{R}{20t^2}ight)$.

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

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