A person can throw a stone to a maximum heigh | vedclass

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(B) The maximum height $H$ attained by a projectile is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$.For maximum height,$	heta = 90^\circ$,so $\sin \

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$2h$

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(B) The maximum height $H$ attained by a projectile is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$.For maximum height,$	heta = 90^\circ$,so $\sin 	heta = 1$.Thus,$H_{	ext{max}} = \frac{v_0^2}{2g} = h$.From this,we get $v_0^2 = 2gh$.The horizontal range $R$ is given by $R = \frac{v_0^2 \sin(2	heta)}{g}$.For maximum horizontal range,$	heta = 45^\circ$,so $\sin(2	heta) = \sin(90^\circ) = 1$.Thus,$R_{	ext{max}} = \frac{v_0^2}{g}$.Substituting $v_0^2 = 2gh$ into the equation for $R_{	ext{max}}$,we get $R_{	ext{max}} = \frac{2gh}{g} = 2h$.

$A$ person can throw a stone to a maximum height of $h$. Determine the maximum horizontal range of the stone in terms of $h$.

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