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(D) The minimum kinetic energy of a projectile occurs at the highest point,where velocity is $v_x = u \cos 	heta$. Thus,$K_{min} = \frac{1}{2} m (u \

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(D) The minimum kinetic energy of a projectile occurs at the highest point,where velocity is $v_x = u \cos 	heta$. Thus,$K_{min} = \frac{1}{2} m (u \cos 	heta)^2$.Given $\frac{K_{min,1}}{K_{min,2}} = \frac{u_1^2 \cos^2 	heta_1}{u_2^2 \cos^2 	heta_2} = 4:1$.The maximum height is $H = \frac{u^2 \sin^2 	heta}{2g}$. Given $\frac{H_1}{H_2} = \frac{u_1^2 \sin^2 	heta_1}{u_2^2 \sin^2 	heta_2} = 4:1$.Dividing the height ratio by the kinetic energy ratio: $\frac{H_1/H_2}{K_{min,1}/K_{min,2}} = \frac{	an^2 	heta_1}{	an^2 	heta_2} = \frac{4}{4} = 1$,so $	an 	heta_1 = 	an 	heta_2$,which means $	heta_1 = 	heta_2$.Since $	heta_1 = 	heta_2$,the ratio of initial velocities is $\frac{u_1^2}{u_2^2} = 4$,so $\frac{u_1}{u_2} = 2$.The range is $R = \frac{u^2 \sin 2	heta}{g}$. The ratio of ranges is $\frac{R_1}{R_2} = \frac{u_1^2}{u_2^2} = 4:1$.

The ratio of minimum kinetic energies of two projectiles of same mass is $4: 1$ and the ratio of maximum heights attained by them is $4: 1$. Then the ratio of their ranges is . . . . . . (in $: 1$)

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