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?

No, unforrtunately you are not on the right track.

First thing is to decide what 't' represents - is it the time for the helicopter in the air, or is it the time the canonball is in the air? It doesn't matter which you pick, as long as you are consistent in how you use 't'. If you pick the canonball, then the two equations of motion for the helicopter and cannonball are:

Helicopter: y_h =1/2 a(t+10)^2, where a = 10m/s^2. Note how the time variable is t+10, because its in the air 10 seconds more than the canonball.

Cannonball: y_c = v_ot - 1/2 gt^2

A the moment of impact y_h = y_c:

So now you have an equation in two unknowns: v_0 and t; we need another equation in order to solve this. That other equation comes from the fact that the cannonball is launched at the smallest velocity possible. Think about what happens if the cannonball goes faster than the minuimum possibe - it rips through the helicoper at a speed greater than the helicopter is risiing. At the minimum possible velocity the cannonball would just touch the heliciopter and then fall away under gravity - in other words at the moment of impact the helicopter and cannonball have the same velocity. This means that:

v_h = at, which equals: v_c = v_0-gt.

Now we have our second equation, so you can solve for v_0.

Try this and see what you get.

I ended up with 198m/s for the minimum speed.

No, unforrtunately you are not on the right track.

First thing is to decide what 't' represents - is it the time for the helicopter in the air, or is it the time the canonball is in the air? It doesn't matter which you pick, as long as you are consistent in how yuo use 't'. If you pick the canonball, then the two equations of motion for the helicopter and cannonball are:

Helicopter: y_h =1/2 a(t+10)^2, where a = 10m/s^2. Note how the time variable is t+10, because its in the air 10 seconds more than the canonball.

Cannonball: y_c = v_ot - 1/2 gt^2

A the moment of impact y_h = y_c:

So now you have an equation in two unknowns: v_0 and t; we need another equation in order to solve this. That other equation comes from the fact that the cannonball is launched at the smallest velocity possible. Think about what happens if the cannonball goes faster than the minuimum possibe - it rips through the helicoper at a speed greater than the helicopter is risiing. At the minimum possible velocity the cannonball would just touch the heliciopter and then fall away under gravity - in other words at the moment of impact the helicopter and cannonball have the same velocity. This means that:

v_h = at, which equals: v_c = v_0-gt.

Now we have our second equation, so you can solve for v_0.

Try this and see what you get.

Would the answer for the minimum speed be 198m/s?