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

$\text {Range}=\frac{u^2 \sin 2 \theta}{g}$<br/><br/> Range for projection angle " $\alpha$ "<br/><br/> $\mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}$<br/><br/> Range for projection angle " $\beta$ "<br/><br/> $$ \begin{aligned} & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \beta}{\mathrm{g}} \\\\ & \alpha+\beta=90^{\circ}(\text { Given }) \\\\ & \Rightarrow \beta=90^{\circ}-\alpha \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin \left(180^{\circ}-2 \alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} \\\\ & \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}=\frac{1}{1} \end{aligned} $$