Two projectiles A and B are thrown with initi | vedclass

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(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$.For projectile $A$: $u_1 = 40\,m/s$,$	heta_1 =

Vedclass · Accepted Answer

$4:9$

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(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$.For projectile $A$: $u_1 = 40\,m/s$,$	heta_1 = 30^{\circ}$.$R_A = \frac{40^2 \sin(2 	imes 30^{\circ})}{g} = \frac{1600 \sin(60^{\circ})}{g} = \frac{1600 	imes \sqrt{3}}{2g} = \frac{800\sqrt{3}}{g}$.For projectile $B$: $u_2 = 60\,m/s$,$	heta_2 = 60^{\circ}$.$R_B = \frac{60^2 \sin(2 	imes 60^{\circ})}{g} = \frac{3600 \sin(120^{\circ})}{g} = \frac{3600 	imes \sqrt{3}}{2g} = \frac{1800\sqrt{3}}{g}$.The ratio of their ranges is $\frac{R_A}{R_B} = \frac{800\sqrt{3}/g}{1800\sqrt{3}/g} = \frac{800}{1800} = \frac{8}{18} = \frac{4}{9}$.

Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m/s$ and $60\,m/s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges is $(g = 10\,m/s^2)$.

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