For a given initial velocity,a projectile is | vedclass

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(B) The formula for the maximum height $H$ of a projectile is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$.For the first angle $	heta_1 = 30^\circ$,

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$3:1$

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(B) The formula for the maximum height $H$ of a projectile is given by $H = \frac{v_0^2 \sin^2 	heta}{2g}$.For the first angle $	heta_1 = 30^\circ$,the height is $H_1 = \frac{v_0^2 \sin^2 30^\circ}{2g}$.For the second angle $	heta_2 = 60^\circ$,the height is $H_2 = \frac{v_0^2 \sin^2 60^\circ}{2g}$.The ratio of the heights is $\frac{H_2}{H_1} = \frac{\sin^2 60^\circ}{\sin^2 30^\circ}$.Substituting the values,$\frac{H_2}{H_1} = \frac{(\sqrt{3}/2)^2}{(1/2)^2} = \frac{3/4}{1/4} = 3$.Therefore,the ratio is $3:1$.

For a given initial velocity,a projectile is fired at angles of $30^\circ$ and $60^\circ$. Find the ratio of their maximum heights.

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