Two stones are thrown vertically up at the same time. The first stone is thrown with an initial velocity of 12.5 m/s from a 12th-floor balcony of a building and hits the ground after 4.6 s. With what initial velocity should the second stone be thrown from a 4th-floor balcony so that it hits the ground at the same time as the first stone? Assume equal height floors, and that in each case the stone is dropped from the same height as the ceiling.

Answer 1

time is the same.

first stone
hf=hi+vi*t-1/2 g t^2
figure the height of reach floor, and simplify the equation in this form
0=hi+vi*t-1/2 g t^2
second stone
or -hi=vi*t-1/2 g t^2

second stone> do the same thing, you should get
- hi*1/3=V*t-1/2 g t^2
or -hi=3V*t-3/2 g t^2
set the first and second equation equal
and do some algebra to get

t(3V-vi)=1/2 gt^2(3-1) and check that (I did it in my head).
YOu know t, vi, g, solve for V

check my head algebra.

Answer 2

I'm really confused by this. What am i supposed to use for t in the first equation if i only know that t going up is t and t going down is 4-t. I can't find two variables with one equation and the first equation is solving for hf.