If a body A of mass M is thrown with velocity | vedclass

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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{v^2 \sin(2	heta)}{g}$.For body $A$,the angle of projection is $	het

Vedclass · Accepted Answer

$1:1$

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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{v^2 \sin(2	heta)}{g}$.For body $A$,the angle of projection is $	heta_A = 30^{\circ}$. Thus,$R_A = \frac{v^2 \sin(2 	imes 30^{\circ})}{g} = \frac{v^2 \sin(60^{\circ})}{g}$.For body $B$,the angle of projection is $	heta_B = 60^{\circ}$. Thus,$R_B = \frac{v^2 \sin(2 	imes 60^{\circ})}{g} = \frac{v^2 \sin(120^{\circ})}{g}$.Since $\sin(120^{\circ}) = \sin(180^{\circ} - 60^{\circ}) = \sin(60^{\circ})$,it follows that $R_A = R_B$.Therefore,the ratio of the horizontal range of $A$ to $B$ is $R_A : R_B = 1:1$.

If a body $A$ of mass $M$ is thrown with velocity $v$ at an angle of $30^{\circ}$ to the horizontal and another body $B$ of the same mass is thrown with the same speed at an angle of $60^{\circ}$ to the horizontal,the ratio of the horizontal range of $A$ to $B$ will be:

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