# Motion equations problem

I have a problem from a practice physics exam:
The specifications for design of a playgroud slide say that a child should gain a speed of no more than 15m/s by sliding down the eqipment. How tall can the slide be?

these are the equations that we'll be given in our exam:
v$$_{x}$$ = v$$_{0x}$$ + a$$_{x}$$t
x = $$\frac{1}{2}$$ (v$$_{0x}$$ + v$$_{x}$$)t
x = v$$_{0x}$$t + $$\frac{1}{2}$$a$$_{x}$$t$$^{2}$$
v$$_{x}$$$$^{2}$$ = v$$_{0x}$$$$^{2}$$ + 2a$$_{x}$$x

although i can work out the answer to the problem i had to look up a new equation because i couldn't work out how to do it with the equations that we were given, so i was wondering if it was possible to do it with one of them, so if you could tell me which one would work the best or if i should just start memorising some new equations, but here's what i did anyway:
v$$_{max}$$ = 15m/s where g = -9.81m/s
v$$_{y}$$$$^{2}$$ = -2g$$\Delta$$y
therefore $$\Delta$$y = $$\frac{v_{y}^{2}}{-2g}$$ = $$\frac{15^{2}}{-9.81\times-2}$$
=11m

Thanks