The body was thrown from a height of 200 meters with an initial speed of 2 m / s. How long did this body fall?

To find the duration of the fall of the taken body, we will use the formula: S = h = V0 * t + g * t ^ 2/2.

The values of the variables: h – the height at which the taken body was located (S = 200 m); V0 – initial speed of the downward throw (V0 = 2 m / s); g – acceleration due to gravity (g ≈ 9.8 m / s2).

Let’s transform the formula into a quadratic equation: g * t ^ 2/2 + V0 * t – h = 0.

4.9t ^ 2 + 2t – 200 = 0.

Let’s calculate the discriminant: D = 2 ^ 2 – 4 * 4.9 * (-200) = 3924.

Determine the fall time (roots of the equation):

t1 = (-2 + √3924) / (2 * 4.9) ≈ 6.19 s (correct).

t2 = (-2 – √3924) / (2 * 4.9) ≈ -6.6 s (incorrect).

Answer: The body fell in 6.19 seconds.