simple problem? acceleration? PLEASE HELP ASAP!

[WillParadigm]

9.Superman is supposed to leap tall buildings in a single bound. Suppose that he obeys the normal laws of physics in this feat. Estimate how fast he should be moving when he leaves the ground so that he just clears the top of Disque Hall. (Hint # 1: how fast would he be moving if he just fell off the top? Hint #2: you have to estimate how tall Disque Hall is.)

Disque Hall is I estimate about 160 feet tall.

with hint #1, I mean, I know g=9.8 m/s^2 but how do I use that to help me figure out how fast he should be going to over leap the building?

Please help me!

*edit, alright, I mean, I know this equation is

d=v(i)t + 1/2at^2

so I have d, which is 160 feet, I have a, which is 9.8 m/s^2, but I converted that into feet to match the building, so a, is now 32.2 ft/s^2. and I know I'm trying to solve for the initial velocity, which has to be great enough so that he can jump over the building, but how do I find t? am I supposed to just guess how long it'll take him?! because then I would know all the variables, but that doesn't seem right

[WillParadigm]

ok reedited, so I know what I'm looking for, but the question is, how do I determine how to find them, what equation am I missing?

[arildno]

1. Find out when superman reaches maximum height (what must his velocity be then?)
2. Use the time found here in your distance equation.
This yields an equation for the initial velocity

[WillParadigm]

ok, I can understand that part for finding time, superman's final height

g = 9.8 m/s² and upward velocity is negative, since the gravitational acceleration, which acts downward, is always positive. so when superman reaches his maximum height, his velocity should be zero. right? So, vf = 0.

vf = vi + at
0 = vi + (9.8 m/s²)t

and that's all I got, do I already start estimating how fast he should be going at vi to get t? but then that doesn't make sense, since you need t in the other equation to get vi, so this is what is confusing me

[arildno]

You have, with correct sign:
0=v_{i}-gt_{f}
whatever initial velocity (v_{i}) is, and whatever the "final" time is.
Hence, you have:
t_{f}=\frac{v_{i}}{g}
Substitute this expression for the final time into the "t"-place in your distance equation.

[WillParadigm]

are you a physics professor/major? because you should be... =)

thank you very much, that makes 100% sense now.