JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 9)
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 :
1 : 1
2 : 1
1 : 2
4 : 1
Explanation
$$
\text {Range}=\frac{u^2 \sin 2 \theta}{g}
$$
Range for projection angle " $\alpha$ "
$$ \mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} $$
Range for projection angle " $\beta$ "
$$ \begin{aligned} & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \beta}{\mathrm{g}} \\\\ & \alpha+\beta=90^{\circ}(\text { Given }) \\\\ & \Rightarrow \beta=90^{\circ}-\alpha \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin \left(180^{\circ}-2 \alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} \\\\ & \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}=\frac{1}{1} \end{aligned} $$
Range for projection angle " $\alpha$ "
$$ \mathrm{R}_1=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} $$
Range for projection angle " $\beta$ "
$$ \begin{aligned} & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \beta}{\mathrm{g}} \\\\ & \alpha+\beta=90^{\circ}(\text { Given }) \\\\ & \Rightarrow \beta=90^{\circ}-\alpha \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin \left(180^{\circ}-2 \alpha\right)}{\mathrm{g}} \\\\ & \mathrm{R}_2=\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}} \\\\ & \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}{\left(\frac{\mathrm{u}^2 \sin 2 \alpha}{\mathrm{g}}\right)}=\frac{1}{1} \end{aligned} $$
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