How does the sign of the acceleration depends on the direction of the distance choosen? I am very confused as to how to determine directions in problems. For example I was doing the following problem

A bomber releases a bomb while flying vertically upwards at a velocity of 1500 ft/sec at a height of 10,000 feet. a) How long after it is released will it take the bomb to reach the ground. b) Immediately after releasing the bomb the bomber flies away horizontally at the rate of 1200 feet/sec. How far away from the point at which the bomb strikes the ground is the plane when the bomb strikes? 

So for this question I decided to choose the downward direction as positive for the distance traveled and so for that to happen I know that my acceleration needs also to be positive. In consequence I had $a= 32$, $v= 32t+1500$ and $s= 16t+1500$. Then I just equated $s(t)$ to $10 000 feet $ to see when would the distance traveled by the bomb would be $10,000 feet $ however this did not give me the correcte answer. I am not sure what I did wrong. Please let me know where I went wrong. Now for the second part I took $1200 feet/sec $ as a velocity and decided to integrate it to find the distance traveled. However in the answer they just multiply 1200 feet by 100 second to find its distance. Why is that? 

$$a=\frac{dv}{dt}=32 \\ \int_0^t \frac{dv}{dt}\ dt = \int_0^t 32\ dt \\ v(t)-v(0)=32t \\ v(t) = 32t+v(0)$$  But $v(0)$ is $1500\ ft/s$ in the negative direction (because the bomb is initially travelling upward with the plane and you chose downward as the positive direction).  So $v(t) = 32t-1500$.  Then integrate again to find $s(t)$.
