Ok, yes we did things much the same way (my working below g=10 for lazy..) what I was trying to say was that knowing Sx=18, Vy=0 doesn't tell you anything at all, and doesn't tell you that the initial velocities are linked, I thought it confused matters and suggested that the way to solve would be something to do with how distance varies over time/whatever. We were linking together with time - the horizontal distance was a constant. I wasn't clear either! the reasons I jumped on 18cos30 for max height are A. Pretty sure it's wrong given both our answers of initial velocity at 20ms (intuitively and I worked it out), and B. The way froggy appeared to do the calculation was to pluck height out of thin air and use it to get velocity, which (given I didn't have any of the specific projectile equations) seemed unlikely to be the way to go. at this point if I was going to calculate height I would be using s=ut + .5at^2 as it seems a bit silly to use a specific projectile equation plucked from the air. We know time very simply as we know V. Interestingly at at this point I thought well actually what if you can pluck the height out without calculating V or t ? The range is fixed at 36m so there is only one path it can possibly follow, for which you don't need to know gravity or initial velocity. You can describe this as a parabola fairly easily. The differential at time 0 and using the angle at take off helps you get the height to drop out. I haven't got this written down legibly! But it gave the same answer which is always good. I will end by saying I haven't done any maths for a long long time so did this as a sanity/mind check, so hey I'm rusty and probably haven't helped.