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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\alpha)}{g}$,where $u$ is the initial velocity,$\alpha$ is

Vedclass · Accepted Answer

$1:1$

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(B) The horizontal range $R$ of a projectile is given by the formula: $R = \frac{u^2 \sin(2\alpha)}{g}$,where $u$ is the initial velocity,$\alpha$ is the angle of projection,and $g$ is the acceleration due to gravity.For the first angle of projection $\alpha_1 = (45^\circ - 	heta)$:$R_1 = \frac{u^2 \sin[2(45^\circ - 	heta)]}{g} = \frac{u^2 \sin(90^\circ - 2	heta)}{g} = \frac{u^2 \cos(2	heta)}{g}$.For the second angle of projection $\alpha_2 = (45^\circ + 	heta)$:$R_2 = \frac{u^2 \sin[2(45^\circ + 	heta)]}{g} = \frac{u^2 \sin(90^\circ + 2	heta)}{g} = \frac{u^2 \cos(2	heta)}{g}$.Comparing the two ranges:$\frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2	heta)}{g}}{\frac{u^2 \cos(2	heta)}{g}} = \frac{1}{1}$.Thus,the ratio of the horizontal ranges is $1:1$.

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of:

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