A high school student shoots a 0.04 kg arrow straight up in the air at 30 m/s during an archery class in the school gym. a) assuming no air resistance, what is the max height that the arrow could reach?

Since this question is titled "work, energy., power..." you should try and solve using the principle of conservation of energy. The kinetic energy that the arrow is initially given is KE = 1/2 mv^2. This KE "bleeds off" as the arrow rises against gravity, in exchange for an increase in potential energy. The amount of PE gain is: PE = mgh. So, set KE = PE, and see what you get for h.

d = \frac{1}{2}gt^2

v = gt

Going straight up, gravity will slow something down at the same rate it would speed it up, if it were dropped. Thinking reversely, we can find out how far it would have fallen when it reaches 30m/s.

Take the second equation and solve for time... this will tell you how long it has fallen. Then plug that into your first equation and you will have a magical distance.

30m/s = 10m/s^{2}t

3s = t

I think you can try the rest.
\frac{30m/s}{10m/s^2} = t

I got 46m

That sounds about right...