G A 10kg weight is suspended from the ceiling by a spring. The weight-spring system is at equilibrium with the bottom of the weight about 1m above the floor. The spring is then stretched until the weight is just above the eggs. When the spring is released, the weight is pulled up by the contracting spring and then falls back down under the influence of gravity. On the way down, it:

Reverses its direction of travel precisely as it reaches the eggs

Explanation:

At the top of the motion: The spring is stretched the least, so, the potential energy of the spring is at a minimum. The mass is lifted as high as it will go, so, the potential energy due to gravity is at a maximum.

When the spring is now at the maximum, now, the maximum potential at the maximum height is

Equal to the energy stored in spring

And the energy stored in spring.

The net force on the object can be described by Hooke’s law, and so the object undergoes simple harmonic motion. Note that the initial position has the vertical displacement at its maximum value X; v is initially zero and then negative as the object moves down; and the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point.

So this apply to the body given, when it get to the egg and it is released the weight will move upward 1m above the equilibrium point and it will return downward 1m below the equilibrium point, that is reverses its direction of travel precisely as it reaches the eggs.

i guess it’s a and b

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.