In an action movie, a helicopter with the heroes on board is seen escaping from an underground bunker inside a volcano. The Volcano chimney is 605m tall. The helicopter is moving up with a constant acceleration of 10m/s^2. the villain realized of this and shoots the helicopter with a cannon. If it took the villain 10s to prepare the cannon and shoot, What is the minimum speed the cannon ball has to have to hit the helicopter?

what is the minimum speed the cannon ball has to have to hit the helicopter?

Assuming the cannon ball is shot at exactly this minimum speed, what is the cannon ball speed when it reaches the helicopter?

The pilot of the helicopter manages to maneuver it safe from the cannon ball. As you expect the cannonball will reach a maxium altitude and fall back down. How long will it take for the ball to return to the bunker?

**This is all I have so far:
t=0; when the cannonball is launched
t of cannonball= -10s
acceleration of helicopter= 10m/s^2
acceleration of cannonball= -9.80m/s^2

**I just don't exactly know which direction to take but these are the equations and solutions I have:
V=at+Vo
0m/s=10m/s^2(-10s)+Voh
Voh=100m/s

Xh=1/2at^2+VohT+Xoh
605m=1/2(10)(-10)^2+(100)(-10)+Xoh
605m=500-1000+Xoh
1105=Xoh

***Am I even on the right track to end up with the answers needed?

I don't quit understand what you were trying to do.

In v = u + at,
v = final velocity,
u = initial velocity,
a = acceleration,
t = time.

From what you tried, you are saying that at t = -10 s, the final velocity of the helicopter (which is accelerating up at 10 m/s^2) is at rest, and you are finding it's initial velocity. That doesn't sound right to me.

You need to think about distance and time here (for the cannonball to reach the helicopter).

In time t seconds, the helicopter will he 605 m high. We can use the equation: h = h_o + u(t + 10) + -0.5a(t+10)^2

ho = initial height (0 m)
h = height (605 m)
u = initial velocity (0 m/s)
t = time elapsed after the villain realise whatever is happening (Note that I added 10 s because the helicopter has an advance of 10 s on the villain and I'm keeping the variables you started with).
a = acceleration (10 m/s^2)

We know that the helicopter must be at rest when it starts, it hasn't yet taken off at t = 0 s, hence why u = 0 m/s. Does it make sense? You can find t.

The next equation is about the cannonball. At time t seconds, the bullet is h m high.

h = h_o + ut + 0.5at^2

ho = initial height (0 m)
h = height (605 m)
u = initial velocity
a = acceleration on ball (-9.80 m/s^2)
t = time (which you got above).

This should give you the first part.