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 $1:1$