Question

If a body $A$ of mass $M$ is thrown with velocity $v$ at an angle of $30^{\circ}$ to the horizontal and another body $B$ of the same mass is thrown with the same speed at an angle of $60^{\circ}$ to the horizontal, the ratio of horizontal range of $A$ to $B$ will be

• A. $1 : 3$
• B. $1 : 1$
• C. $1 : \sqrt 3 $
• D. $ \sqrt 3 : 1$

Answer 1

Answer: (B)

For the given velocity of projection $u$, the horizontal range is the same for the angle of projection $\theta$ and $90^{\circ}-\theta$
Horizontal range $R=\frac{u^{2} \sin 2 \theta}{g}$
$\therefore $ For body $A R_{A}=\frac{u^{2} \sin \left(2 \times 30^{\circ}\right)}{g}=\frac{u^{2} \sin 60^{\circ}}{g}$
For body $B R_{B}=\frac{u^{2} \sin \left(2 \times 60^{\circ}\right)}{g}$
$R_{B}=\frac{u^{2} \sin 120^{\circ}}{g}=\frac{u^{2} \sin \left(180^{\circ}-60^{\circ}\right)}{g}=\frac{u^{2} \sin 60^{\circ}}{g}$
The range is the same whether the angle is $\theta$ or $90^{\circ}-\theta$.
$\therefore $ The ratio of ranges is $1: 1$