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(B) The formula for the horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.Given $R = 20 \ m$ and $	heta = 30^{\circ}$,we have $20 = \

Vedclass · Accepted Answer

$\frac{5}{\sqrt{3}}$

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(B) The formula for the horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.Given $R = 20 \ m$ and $	heta = 30^{\circ}$,we have $20 = \frac{u^2 \sin(60^{\circ})}{g}$.Thus,$\frac{u^2}{g} = \frac{20}{\sin 60^{\circ}} = \frac{20}{\sqrt{3}/2} = \frac{40}{\sqrt{3}}$.The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 	heta}{2g}$.Substituting the values,$H = \frac{1}{2} 	imes \left(\frac{u^2}{g}ight) 	imes \sin^2 30^{\circ}$.$H = \frac{1}{2} 	imes \frac{40}{\sqrt{3}} 	imes \left(\frac{1}{2}ight)^2$.$H = \frac{20}{\sqrt{3}} 	imes \frac{1}{4} = \frac{5}{\sqrt{3}} \ m$.

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$,the maximum height reached by it is (in meters):

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