A stone projected with a velocity u at an ang | vedclass

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Question

(a)${H_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and ${H_2} = \frac{{{u^2}{{\sin }^2}(90 - 	heta )}}{{2g}} = \frac{{{u^2}{{\cos }^2}	heta }}{{2g}

Vedclass · Accepted Answer

$R = 4\sqrt {{H_1}{H_2}} $

Vedclass · Answer

(a)${H_1} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$ and ${H_2} = \frac{{{u^2}{{\sin }^2}(90 - 	heta )}}{{2g}} = \frac{{{u^2}{{\cos }^2}	heta }}{{2g}}$

${H_1}{H_2} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} 	imes \frac{{{u^2}{{\cos }^2}	heta }}{{2g}} = \frac{{{{({u^2}\sin 2	heta )}^2}}}{{16{g^2}}} = \frac{{{R^2}}}{{16}}$

$R = 4\sqrt {{H_1}{H_2}} $

Column $-I$ Angle of projection	Column $-II$
$A.$ $\theta \, = \,{45^o}$	$1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$
$B.$ $\theta \, = \,{60^o}$	$2.$ $\frac{{g{T^2}}}{R} = 8$
$C.$ $\theta \, = \,{30^o}$	$3.$ $\frac{R}{H} = 4\sqrt 3 $
$D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$	$4.$ $\frac{R}{H} = 4$

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