A projectile is fired from horizontal ground | vedclass

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(B) The initial range of the projectile is given by $d = \frac{v^2 \sin 2	heta}{g}$.At the highest point,the vertical velocity is zero and the horizo

Vedclass · Accepted Answer

$0.95$

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(B) The initial range of the projectile is given by $d = \frac{v^2 \sin 2	heta}{g}$.At the highest point,the vertical velocity is zero and the horizontal velocity is $u = v \cos 	heta$. The maximum height reached is $H_{\max} = \frac{v^2 \sin^2 	heta}{2g}$.After entering the new region,the projectile falls from height $H_{\max}$ under gravity $g^{\prime} = \frac{g}{0.81}$. Let $t$ be the time taken to reach the ground from the highest point.Using $H_{\max} = \frac{1}{2} g^{\prime} t^2$,we get $t = \sqrt{\frac{2 H_{\max}}{g^{\prime}}} = \sqrt{\frac{2 (v^2 \sin^2 	heta / 2g)}{g / 0.81}} = \sqrt{\frac{v^2 \sin^2 	heta 	imes 0.81}{g^2}} = \frac{0.9 v \sin 	heta}{g}$.The horizontal distance covered in this time is $d_1 = u 	imes t = (v \cos 	heta) 	imes \left( \frac{0.9 v \sin 	heta}{g} ight) = \frac{0.9 v^2 \sin 	heta \cos 	heta}{g} = \frac{0.45 v^2 \sin 2	heta}{g}$.The total new range is $d^{\prime} = \frac{d}{2} + d_1 = \frac{v^2 \sin 2	heta}{2g} + \frac{0.9 v^2 \sin 2	heta}{2g} = \frac{v^2 \sin 2	heta}{2g} (1 + 0.9) = \frac{v^2 \sin 2	heta}{2g} (1.9) = 0.95 \left( \frac{v^2 \sin 2	heta}{g} ight) = 0.95 d$.Thus,$n = 0.95$.

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