A small bag is released from a helicopter that is climbing steadily at 1.5 m/s. after 2 seconds,

(a) find the velocity of the mail bag?

(b) how far is it below the helicopter?

There is not enough information. When exactly was the bag released? How high, or for how much time has the helicopter been rising? If it's after a short while, the bag may hit the ground before 2 seconds...

It douse not hit the ground (a) should be 20

It was released the question is for 2 seconds after it was released

Ok, use the formula v=u+at

v is the speed of the bag
u the velocity of the helicopter when the bag was released
a the acceleration due to gravity
t the time elapsed.

So would (a) be 21.5 m/s?
Do u know (b) or how would I get it

a) Nope. You need to take into consideration the directions as well. Ok, let's take the downwards direction as positive. The initial velocity is therefore -1.5m/s, the acceleration +10m/s^2 and the time is 2 s.

b) Use the formula : s = ut + \frac12 at^2

Here, s is the displacement, from the point of release and the rest of the letters are the same as previously.

Find the displacement 's'.

Then, find the distance covered by the helicopter in 2 s during its ascension.

Add the two distances to get the separation between the two.