[...] lolthx Chris. I've written the Horizontal Accel as 0, not as 1, so i'll re do it from there now we use S=ut+1/2at^2 again S=xt+1/2t^2 (as a=1) 30=2.47x + 1/2*2.47^2 x=10.91m/s x(Initial Speed) to 2sf=11m/s Lol thanks Chris, epic fail on my part there

I know what you mean, yeah. But do you know what i mean? :/ If he's throwing it at an angle upwards, we don't know the initial horizontal or vertical speed. We don't know the time, we only know the horizontal distance (for uneven ground) and the vertical acceleration. If you do a quick diagram yourself, i think you might see what i mean. Whether it's vcos34 or vsign34, it's still a proportion of a variable that we don't know. So basically, we know we know horizontal accel is 0, horizontal displacement is 30.

horizontally: 30 = utcos(34) vertically: v = usin(34) + gt Using the equation derived from the horizontal motion we see: v = usin(34) + 30g/(ucos(34)) (v^2 = (usin34)^2 + 900g^2/(ucos34)^2 + 60gtan34) using v^2 = u^2 + 2as we see: v^2 = (usin34)^2 + 60g) hence 60g = 60gtan(34) + 900g^2/(ucos34)^2 1-tan34 = 15g/(ucos34)^2 u=25.6ms^-1 this might be wrong i cba to check

Hard via text, i know, but: - Lets work out the vertical, so we know the time it takes to hit the ground - S-30m - U-0 - V-? - A-9.8 - T-? Why have you started with initial vertical velocity as 0? The basketball starts with speed s at 53 degrees above the horizontal, so the initial vertical velocity is s * sin 53, and we don't know what s is. Also in your edit, why is a=1? horizontal acceleration is 0, vertical acceleration is 9.18 --- Also, i'm gonna go back on myself on that one, i think the horizontal range in this case means when it hits the floor at the bottom of the cliff. There is a formula for uneven ground but you're not gonna like it: http://en.wikipedia.org/wiki/Range_of_a_projectile#Uneven_Ground Believe me, sticking values into that gives you something like 3.5 million on one side of the equation, and v^4, v^2 and v on the other side.