(a) what is the bag's velocity?
m/s
(b) how far has the bag fallen?
m
(c) how far below the helicopter is the bag?
2006-10-15 13:38:04 · 3 answers · asked by hi 2 in Science & Mathematics ➔ Physics
INITIAL VELOCITY = - 5m/s [taking downward direction positive]. Dist covered =-5*5 + 1/2 *g *25 =100m [g=10]
BY THIS TIME HELICOPTER COVERED 5*5=25m [considering it is moving with uniform velocity]
SO BAG IS 125m BELOW HELICOPTER.
2006-10-15 21:14:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I disagree with Dr Jones. Ignores the original velocity 5.0 m/s that the bag starts with.
Besides: I don't have a problem with most theoretical questions when they ignore wind resistance, friction etc. But come on! A bag dropped? From a helicopter with all the wash from the props?
2006-10-15 15:13:53 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
a) g=9.8m/s therefore 48m/s (9.8x5)
b)147m (9.8 +2x9.8 +3x9.8 etc)
c)172m (helicopter has risen 25m, bag fell 147, 25+147=172)
next time do ur own hmwk! lol
2006-10-15 13:50:28 · answer #3 · answered by Dr Jones 2 · 0⤊ 0⤋