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Physics Question on Motion in a plane

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 : 3\sqrt 3

D

3:1 \sqrt 3 : 1

Answer

1 : 1

Explanation

Solution

For the given velocity of projection u, the horizontal range is the same for the angle of projection θ \theta and 90∘−θ90^\circ - \theta
Horizontal range R = u2 sin⁡ 2θg\frac{ u^2 \ \sin \ 2 \theta }{ g}
∴\therefore For body A RA=u2 sin⁡(2×30∘)g=u2 sin⁡60∘g R_A = \frac{ u^2 \ \sin ( 2 \times 30^\circ) }{g} = \frac{u^2 \ \sin 60^\circ }{ g}
For body B RB=u2 sin⁡(2×30∘)g=u2 sin(2×60∘)gR_B = \frac{u^2 \ \sin ( 2 \times 30^\circ) }{ g} = \frac{ u^2 \ sin ( 2 \times 60^\circ)}{ g}
RB=u2 sin 120∘g=u2 sin⁡(180∘−60∘)g=u2 sin⁡60∘gR_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 θ \theta or 90∘−θ90^\circ - \theta.
∴\therefore The ratio of ranges is 1 : 1