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While rising vertically at 10 m/s, a helicopter passenger falls out attached to a rope. How much rope has been pulled after 2 seconds?

2007-01-15 12:28:46 · 7 answers · asked by toucan248 1 in Science & Mathematics Physics

7 answers

use the formula

d= 1/2at^2 + vt

here t = 2s, v = 10m/s, and a= 9.8m/s^2

d = 1/2x9.8x4 + 10x2 = 39.6m

2007-01-15 12:37:42 · answer #1 · answered by catarthur 6 · 0 0

OK...
the moment he fall his speed was 10 m/s vertically up.
but his acceleration downward is 10 m/s^2...
with Newton's equations.
d=V.t+(0.5) a t^2 => d=(10)(2)+(0.5)(-10)(2)(2)
= 20-20=0 meters
this is the distance that he fall from the point he started falling.
but the helicopter rose from that point d=(10)(2)=20meters
then the rope has been pulled 20 meters in 2 seconds

if you want any other questions about that send to
(jemi1000000@yahoo.com)

2007-01-16 05:58:19 · answer #2 · answered by JwH 2 · 0 0

20 meters, because for every second the helicopter goes up 10 meters so after 2 seonds it would be 20 meters of rope, and use the formula d=rt
10m/s * 2seconds = 20 m

The units cancel out too.

2007-01-15 20:42:19 · answer #3 · answered by tallwhiteguy00 2 · 0 0

If there were no upward velocity, the passenger will fall through a distance of

0.5 g tt = 19.6 m.

The length of the rope pulled will be 19.6 m.

Since both the passenger and helicopter has an initial upward speed of 10m/s,

The length of rope pulled will be the same as 19.6m
------------------------------------------------------------------------------------------------
OR

Initial velocity of the passenger is minus 10m/s (upward is taken as minus).

Acceleration is 9.8m/s^2 down ward.

In 2 second the velocity changes by 2 x 9.8 = 19.6m/s.

Final velocity is (minus 10) + 19.6 = 9.6m/s.

Distance traveled is (average velocity x time)

Average velocity is (minus 10 + 9.6)/ 2 = minus0.2 m/s.

Distance = minus 0.2 x 2 = 0.4 m.

By this time the helicopter has moved through a distance of
-10x2 = -20m

Distance between the passenger and helicopter is (-0.4)- (- 20 )
= 19.6m
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Or

Using S = ut + 0.5 a t^2

S = -10 x 2 + 0.5 x 9.8 x 2 x 2

S = minus 0.4 m

The passenger is 0.4 m up from the initial position.

By this time the helicopter has moved through a distance of 10x2 = 20m.

Therefore, the length of the rope pulled is 20 - 0.4 = 19.6 m

_------------------------------------------------------------------------------------------------

2007-01-15 21:45:05 · answer #4 · answered by Pearlsawme 7 · 0 0

Twenty meters. It seems so obvious that I think I must be missing something vital.

2007-01-15 20:32:47 · answer #5 · answered by robertspraguejr 4 · 0 0

just tell your teacher this:

you went to college. tell me the answer

lol jk

2007-01-15 20:32:04 · answer #6 · answered by Anonymous · 0 0

enough to hang them?

2007-01-15 20:31:31 · answer #7 · answered by Yahoo Answer Rat 5 · 0 0

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