MCQ
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 :
  • A
    $4: 1$
  • B
    $2: 1$
  • C
    $1: 2$
  • $1: 1$

Answer

Correct option: D.
$1: 1$
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 " $\beta$ "

$R _2=\frac{ u ^2 \sin 2 \beta}{ g }$

$\alpha+\beta=90^{\circ}(\text { Given })$

$\Rightarrow \beta=90^{\circ}-\alpha$

$R _2=\frac{ u ^2 \sin 2\left(90^{\circ}-\alpha\right)}{ g }$

$R _2=\frac{ u ^2 \sin \left(180^{\circ}-2 \alpha\right)}{ g }$

$R _2=\frac{ u ^2 \sin 2 \alpha}{ g }$

$\Rightarrow \frac{ R _1}{ R _2}=\frac{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}{\left(\frac{ u ^2 \sin 2 \alpha}{ g }\right)}=\frac{1}{1}$

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