These types of problems can all be solved by knowing the relationship between the position, velocity, and acceleration equations. In the following, by taking the derivative you can move from one equation to the next: $ \text{position} \to \text{velocity} \to \text{acceleration} $ Similarly, to go from one equation to the next below, you need to integrate the relevant equation: $ \text{acceleration}\to\text{velocity}\to\text{position} $ These equations make more sense when you know the general forms for each equation (assuming constant linear acceleration). The general position equation is given by $ s(t) = \frac{1}{2}at^2+v_0t+s_0 $ where $a$ is acceleration, $t$ is time, $v_0$ is initially velocity, and $s_0$ is the initial position. The general equation for velocity is $ v(t) = at+v_0 $ using the same variables. Finally, the equation for constant linear acceleration is simply $ a(t)=a $
Here you're given the height, or position, equation and you want to get the velocity. From what I wrote above, it's clear that you want to take the derivative to get the velocity.