A body is projected from the ground at an ang | vedclass

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(D) For projectile motion, the angle of projection is $	heta = 	an^{-1}(\sqrt{7})$, so $	an 	heta = \sqrt{7}$.At any height $y$, the vertical comp

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$\frac{3}{4}$

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(D) For projectile motion, the angle of projection is $	heta = 	an^{-1}(\sqrt{7})$, so $	an 	heta = \sqrt{7}$.At any height $y$, the vertical component of velocity is $v_y^2 = u_y^2 - 2gy$.At $y = \frac{H}{2}$, where $H = \frac{u^2 \sin^2 	heta}{2g}$ is the maximum height:$v_y^2 = u^2 \sin^2 	heta - 2g \left( \frac{u^2 \sin^2 	heta}{4g} ight) = u^2 \sin^2 	heta - \frac{u^2 \sin^2 	heta}{2} = \frac{u^2 \sin^2 	heta}{2}$.The horizontal component of velocity remains constant: $v_x = u \cos 	heta$.The speed $v$ at $y = \frac{H}{2}$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 \cos^2 	heta + \frac{u^2 \sin^2 	heta}{2}}$.Given $v = nu$, we have $n^2 u^2 = u^2 \cos^2 	heta + \frac{u^2 \sin^2 	heta}{2}$.$n^2 = \cos^2 	heta + \frac{\sin^2 	heta}{2} = \cos^2 	heta + \frac{1 - \cos^2 	heta}{2} = \frac{1 + \cos^2 	heta}{2}$.Since $	an 	heta = \sqrt{7}$, $	an^2 	heta = 7$, so $\sec^2 	heta = 1 + 7 = 8$, which means $\cos^2 	heta = \frac{1}{8}$.$n^2 = \frac{1 + 1/8}{2} = \frac{9/8}{2} = \frac{9}{16}$.Therefore, $n = \frac{3}{4}$.

$A$ body is projected from the ground at an angle of $\tan ^{-1}(\sqrt{7})$ with the horizontal. At half of the maximum height, the speed of the body is '$n$' times the speed of projection. The value of '$n$' is

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