A ball is thrown straight up from the ground

with an initial velocity of 43.5 m/s; at the
same instant, a ball is dropped from the roof
of a building 15.4 m high.
After how long will the balls be at the
same height? The acceleration of gravity is
10 m/s^2
Answer in units of s.

the part about acceleration of gravity throws me off :(

1 answer

just think of the equations of motion. You want the heights to be the same, so

0 + 43.5t - g/2 t^2 = 15.4 - g/2 t^2

The actual value of g does not matter, since both balls are affected by it. So, you wind up with

43.5t = 15.4

see the heights at

http://www.wolframalpha.com/input/?i=0+%2B+43.5t+-+5+t^2+%3D+15.4+-+5+t^2