Physics problem, Please help.?

Question

A ball is dropped from the top of a 40 m high building. At what INITIAL VELOCITY must a second ball be throw from the top of the building 2.00 s later, such that both balls arrive at the ground at the same time??

Givens...
delta y = -40m
initial velocity of 1st ball = 0 m/s (dropped)

Needs...
initial velocity of second ball = ???


I have no idea on how to do this?!?! i think i have to make two equations equal to each other but i dont think i can because the second ball came 2 seconds later. Please help.



delta x = (Vix)(t)
Vix= Vi cos(pheta
Vy(final velocity) = Viy -gt
delta y = viyt - 1/2gt^2
Vy^2 = Viy^2-2g(delta y)
Viy = Vi sin(pheta

Answer 1

answer is approxomately 42.476m/s

only one formula here
S=ut + (at^2)/2
s- distance travelled = 40m
u- initial valocity
a - acc due to gravity-9.8m/s^2
t- time
in the first case the initial velocity is 0m/s^2
so
40=(9.8*t^2)/2
t=2.857 sec
as both the balls have to touch the ground at same time and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m.
so
40=u*0.857+(9.8*(0.857^2))/2
solving u=42.476m/s
the initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.
that is pheta=0 in the formula given un the question.

Answer 2

Solve for the time to fall. 1/2gt^2 = D

Use that time minus the 2 seconds to find the v0 for the second ball. 1/2gt^2+v0*t = D

Answer 3

You can set the equations equal because they fall the SAME HEIGHT.

Answer 4

U know that "X" and "g" are constants.

It's just the hint.....