Motion In Two Dimensions Question 2
Question 2 - 25 January - Shift 2
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 :
(1) $4: 1$
(2) $2: 1$
(3) $1: 2$
(4) $1: 1$
Show Answer
Answer: (4)
Solution:
Formula: Horizontal range
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 " $\beta$ "
$R_2=\frac{u^{2} \sin 2 \beta}{g}$
$\alpha+\beta=90^{\circ}$ (Given)
$\Rightarrow \beta=90^{\circ}-\alpha$
$R_2=\frac{u^{2} \sin 2(90^{\circ}-\alpha)}{g}$
$R_2=\frac{u^{2} \sin (180^{\circ}-2 \alpha)}{g}$
$R_2=\frac{u^{2} \sin 2 \alpha}{g}$
$\Rightarrow \frac{R_1}{R_2}=\frac{(\frac{u^{2} \sin 2 \alpha}{g})}{(\frac{u^{2} \sin 2 \alpha}{g})}=\frac{1}{1}$
