Ball, gravity, velocity problem

[rcmango]

having trouble setting this up and isolating the key information to solve this problem. here it is.

question: A ball is thrown straight up from the edge of the roof of a building that is 40.0 m high, and a second ball is dropped from the roof 2.00 s later. What is the initial velocity of the first ball if both balls hit the ground at the same time?

answer: 15.6 m/s

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my effort: alright i started by trying to apply these formulas..

v0 = v + at
...
x = x0 + v0t + 1/2at^2 --> tried using this one.
...
v^2 = v0 + 2a(change x)

i think I'm not really using x in the equations so i'll change them to y because were doing vertical drops here.

...

also, i must be forgetting steps here.
i thought that the middle formula would work best, so i plugged in what i knew
and used gravity (9.80)

what we know:
x = 0 (because x final hits the ground)
x0 = 40.0m because that's where both balls are when there first released.
v0 = that's the variable were trying to find
t = 2seconds ? NOT SURE if we can use 2 seconds here.
a = i used gravity for this variable (9.80) I'm not sure why we use - or + though, maybe a quick explanation here please.

...

okay. so as i said, i used the 2nd equation: x = x0 + v0t + 1/2at^2
i plugged the answers in: and calculated: 10.2 m/s

i think I'm missing a big step here.

please help. thanks.

 

[OlderDan]

I don't know if you are using the correct equations. There are mistakes in the way you wrote them here. Fix the mistakes in what you wrote, and make sure you are using the correct equations.

The time of flight for the first ball is two seconds more than the time of flight of the second ball. If up is positive, the final displacement is -40m for each.

 

[sullyhet]

where did that answer come from, are you sure that that's the right answer, if it is thrown up and hits the ground at the same time that another ball is just simply droped then that means that the first ball moves up and then down within 2 seconds and then in that case it must be around 9.8m/s for the inital velocity of the first ball in order for it to travel up for one second and then back down for one second in order to match the time of impact with the ground of ball 2

i could be wrong but i think you have got a trick question bud