Question

Two objects are projected with same velocity ' $u$ ' however at different angles $\propto$ 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 :

• A. $1:2$
• B. $4: 1$
• C. $2: 1$
• D. $1:1$

Answer 1

Answer: (D)

$\text { Range }=\frac{ u ^2 \sin 2 \theta}{ g }$
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}$