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

The Correct Option is 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 ( 2 \times 30^\circ) }{g} = \frac{u^2 \ \sin 60^\circ }{ g}$
For body B $ R_B = \frac{u^2 \ \sin ( 2 \times 30^\circ) }{ g} = \frac{ u^2 \ sin ( 2 \times 60^\circ)}{ g} $
$ R_B = \frac{u^2 \,sin \ 120^\circ}{ g} = \frac{ u^2 \ \sin (180^\circ - 60^\circ )}{ 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