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}} $

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