In this problem, we analyze the motion of a 6-kilogram object thrown downward from an initial height of 20 meters, ultimately reaching a final speed of 30 meters per second just before impact with the ground. To solve for the initial speed, we apply the principle of conservation of energy, which states that the total mechanical energy (kinetic plus potential) remains constant in the absence of non-conservative forces.

We start by defining our energy terms. The initial kinetic energy (K_initial) is given by the formula:

K_initial = \frac{1}{2} m v_{initial}^2

where m is the mass of the object and v_initial is the initial velocity we want to find. The initial gravitational potential energy (U_initial) is calculated as:

U_initial = m g y_{initial}

where g is the acceleration due to gravity (approximately 9.8 m/s²) and y_initial is the height (20 meters).

At the final state, just before hitting the ground, the kinetic energy (K_final) is:

K_final = \frac{1}{2} m v_{final}^2

and the potential energy (U_final) is zero since the height is 0 meters.

Setting up the conservation of energy equation:

K_initial + U_initial = K_final + U_final

Substituting the expressions for kinetic and potential energy, we have:

\frac{1}{2} m v_{initial}^2 + m g y_{initial} = \frac{1}{2} m v_{final}^2

Since mass (m) appears in all terms, it can be canceled out, simplifying our equation to:

\frac{1}{2} v_{initial}^2 + g y_{initial} = \frac{1}{2} v_{final}^2

To eliminate fractions, we can multiply the entire equation by 2:

v_{initial}^2 + 2 g y_{initial} = v_{final}^2

Rearranging to solve for the initial velocity squared gives us:

v_{initial}^2 = v_{final}^2 - 2 g y_{initial}

Substituting the known values (v_final = -30 m/s, g = 9.8 m/s², and y_initial = 20 m), we find:

v_{initial}^2 = (-30)^2 - 2 \cdot 9.8 \cdot 20

Calculating this yields:

v_{initial}^2 = 900 - 392 = 508

Taking the square root gives us:

v_{initial} = \pm \sqrt{508} \approx \pm 22.5 \text{ m/s}

Since the object is thrown downward, we take the negative value, resulting in:

v_{initial} \approx -22.5 \text{ m/s}

This indicates that the object was thrown downward with an initial speed of approximately 22.5 meters per second, confirming that both upward and downward throws could yield the same final speed, but only the downward throw aligns with the problem's conditions.