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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 $2^{\text {nd }}$ object will be :
$4: 1$
$2: 1$
$1: 2$
$1: 1$
Solution
$\text { Range }=\frac{ u ^2 \sin 2 \theta}{ g }$
Range for projection angle " $\alpha$ "
$R _1=\frac{ u ^2 \sin 2 \alpha}{ g }$
Range for projection angle " $\beta$ "
$R _2=\frac{ u ^2 \sin 2 \beta}{ g }$
$\alpha+\beta=90^{\circ}(\text { Given })$
$\Rightarrow \beta=90^{\circ}-\alpha$
$R _2=\frac{ u ^2 \sin 2\left(90^{\circ}-\alpha\right)}{ g }$
$R _2=\frac{ u ^2 \sin \left(180^{\circ}-2 \alpha\right)}{ g }$
$R _2=\frac{ u ^2 \sin 2 \alpha}{ g }$
$\Rightarrow \frac{ R _1}{ R _2}=\frac{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}=\frac{1}{1}$
