$ for a projectile is given by:

$$ H = \frac{u^2 \sin^2(\theta)}{2g} $$

Here:

$$u$$ is the initial velocity
$$\theta$$ is the angle of projection
$$g$$ is the acceleration due to gravity

For the first case:

Initial velocity, $$u = 10 \text{ m/s}$$
Angle of projection, $$\theta = 30^\circ$$

Substituting these values into the formula:

$$ H_1 = \frac{(10)^2 \sin^2(30^\circ)}{2 \times 9.8} $$

We know that $$ \sin(30^\circ) = \frac{1}{2} $$, so:

$$ H_1 = \frac{100 \left( \frac{1}{2} \right)^2}{2 \times 9.8} = \frac{100 \times \frac{1}{4}}{2 \times 9.8} = \frac{25}{19.6} $$

For the second case, the mass and the angle of projection are doubled, but the mass does not affect the height. Therefore, the new angle of projection is:

$$ \theta' = 2 \times 30^\circ = 60^\circ $$

Using the same initial velocity $$u = 10 \text{ m/s}$$, we substitute into the formula:

$$ H_2 = \frac{(10)^2 \sin^2(60^\circ)}{2 \times 9.8} $$

We know that $$ \sin(60^\circ) = \frac{\sqrt{3}}{2} $$, so:

$$ H_2 = \frac{100 \left( \frac{\sqrt{3}}{2} \right)^2}{2 \times 9.8} = \frac{100 \times \frac{3}{4}}{2 \times 9.8} = \frac{75}{19.6} $$

Now, to find the ratio of the maximum heights, we calculate:

$$ \text{Ratio} = \frac{H_1}{H_2} = \frac{\frac{25}{19.6}}{\frac{75}{19.6}} = \frac{25}{75} = \frac{1}{3} $$

Therefore, the ratio of the maximum height attained in the former to the latter case is 1 : 3, which corresponds to Option C.

Option C: 1 : 3

Circular Motion

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