A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is _____ m. Use g = 10 m s$$^{-2}$$

Let A be at distance $$s$$ from the starting point. The body covers 80 m from A to B in 2 seconds.

Distance from start to A: $$s = \frac{1}{2}gt_1^2$$ where $$t_1$$ is time to reach A.

Distance from start to B: $$s + 80 = \frac{1}{2}g(t_1 + 2)^2$$

$$80 = \frac{1}{2}g[(t_1+2)^2 - t_1^2] = \frac{1}{2}(10)(4t_1 + 4) = 5(4t_1 + 4) = 20t_1 + 20$$

$$20t_1 = 60$$

$$t_1 = 3$$ s

$$s = \frac{1}{2}(10)(9) = 45$$ m

The answer is $$\boxed{45}$$ m.