Ok, basic calculus lesson.
Velocity is the rate of change of position with respect to time meaning ##v(t) = \frac{d}{dt}P(t)##
Acceleration is the rate of change of velocity with respect to time meaning ##a(t)=\frac{d}{dt}v(t) = \frac{d^2}{dt^2}P(t)##
You can reverse these processes by calculating the antiderivative (basically the derivative operation but in reverse, it differs a little for more complex functions)
If you have an acceleration, in this case g, you can write that as a vector. Which way does gravity pull on objects in terms of unit vectors? (or whatevery notation you prefer) <x,y,z> ##x\hat{i} + y\hat{j} + z\hat{k}##
When you calculate the antiderivative, you end up with a constant of integration (when you take the derivative of a constant, you get 0, so you have to add a constant into your function) and you need to use boundery conditions to solve for that constant. So when you do this problem, you start with the acceleration vector, and you move to the velocity vector by way of antiderivatives. You take some kind of condition that you know the value of the velocity for and use that to solve for your constant. So in this case you know the initial velocity, that's how fast sir hat wearer jumps. The cool thing about this problem is that you know the initial velocity at t=0, or the beginning of your motion you're trying to model. So ##v(t) = \int a(t)dt \ \ \text{&}\ \ v(0) = v_0 \text{(initial velocity)}## See if you can calculate this. Post back what you have.
If you need a refresher, I personally like Paul's notes:
http://tutorial.math.lamar.edu/Classes/CalcI/IndefiniteIntegrals.aspx
Also, this might be a better page, but ultimately I don't really know what you know/don't know, so you might be the best person to gather resources.
http://tutorial.math.lamar.edu/Classes/CalcIII/Velocity_Acceleration.aspx