The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)