18 or 72$^o$.
Work Step by Step
Use the level horizontal range formula that is derived in the text, with a range of 2.5 meters. You will find $sin(2 \theta) = 0.5799$ which can be solved for the given angles. These two angles give the same range. The 18$^o$ solution results in a fast, low trajectory, while the higher angle results in the water spending more time in the air, and arcing down onto the target.