Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 1.2s after passing his window. Roger's room is on the fifth floor, 25m above the sidewalk.
How fast are the balloons traveling when they pass Roger's window? Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is 5.0 m high.

Question

Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 1.2s after passing his window. Roger's room is on the fifth floor, 25m above the sidewalk. 
How fast are the balloons traveling when they pass Roger's window? Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is 5.0 m high.

anonymous · Answer

s = v t - (1/2)(g)t^2     
where s is distance from Roger's window to street below
 and v is velocity as it passes his floor. Downward is negative.

-25 = v (1.2) - 0.5 (9.8)(1.2)^2

-v = (25 - 7.056) / 1.2 = 15 m/s

it reached this velocity from gravity 
losing potential energy m g h' 
to gain kinetic energy  (1/2) m v^2 from where it was dropped h' above him 
so (1/2) v^2 = g h'
h' = (v^2)/2 g = (-15)^2 / 2 (9.8) = approx  (225)/(20) = 11.25 m 
or about 2 floors above him.