Two bodies are projected from ground with same speeds 40 m s$$^{-1}$$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60°, with horizontal then sum of the maximum heights, attained by the two projectiles, is ______ m. (Given $$g = 10$$ m s$$^{-2}$$)

Two projectiles launched with the same speed $$u = 40$$ m/s have the same range. One is at $$60°$$.

For same range with same speed, the angles are complementary: $$\theta_2 = 90° - 60° = 30°$$.

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

For $$\theta_1 = 60°$$:

$$H_1 = \frac{40^2 \times \sin^2 60°}{2 \times 10} = \frac{1600 \times \frac{3}{4}}{20} = \frac{1200}{20} = 60 \text{ m}$$

For $$\theta_2 = 30°$$:

$$H_2 = \frac{40^2 \times \sin^2 30°}{2 \times 10} = \frac{1600 \times \frac{1}{4}}{20} = \frac{400}{20} = 20 \text{ m}$$

$$H_1 + H_2 = 60 + 20 = 80 \text{ m}$$

The answer is $$\boxed{80}$$ m.