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(D) The formula for the horizontal range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For the first object projected at angle $\alpha$

Vedclass · Accepted Answer

$1:1$

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(D) The formula for the horizontal range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For the first object projected at angle $\alpha$,the range is $R_1 = \frac{u^2 \sin 2\alpha}{g}$.For the second object projected at angle $\beta$,the range is $R_2 = \frac{u^2 \sin 2\beta}{g}$.Given that $\alpha + \beta = 90^{\circ}$,we have $\beta = 90^{\circ} - \alpha$.Substituting $\beta$ in the expression for $R_2$:$R_2 = \frac{u^2 \sin 2(90^{\circ} - \alpha)}{g} = \frac{u^2 \sin(180^{\circ} - 2\alpha)}{g}$.Since $\sin(180^{\circ} - 	heta) = \sin 	heta$,we get $R_2 = \frac{u^2 \sin 2\alpha}{g}$.Therefore,the ratio is $\frac{R_1}{R_2} = \frac{\frac{u^2 \sin 2\alpha}{g}}{\frac{u^2 \sin 2\alpha}{g}} = \frac{1}{1}$.

Two objects are projected with the same velocity '$u$' but at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^{\circ}$,the ratio of the horizontal range of the first object to the $2^{\text{nd}}$ object will be:

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