Asked by Tay
A helicopter is rising at 4.7 m/s when a bag is dropped from it. (Assume that the positive direction is upward.)
(a) After 2.0 s, what is the bag's velocity?
(b) How far has the bag fallen?
(c) How far below the helicopter is the bag?
(a) After 2.0 s, what is the bag's velocity?
(b) How far has the bag fallen?
(c) How far below the helicopter is the bag?
Answers
Answered by
bobpursley
v(2)=vo*2-9.8(2^2)
see if you can do the rest.
see if you can do the rest.
Answered by
Tay
i still cant do it
Answered by
bobpursley
b) d(2)=Vo*2-1/2 g (4)
oops, I did a error in a)
v(2)=vo -9.8*2 and vo is 4.7
c) Ok, on how far, it fell b), but the helo went up
farbelow= 4.7*2+answerinb)
oops, I did a error in a)
v(2)=vo -9.8*2 and vo is 4.7
c) Ok, on how far, it fell b), but the helo went up
farbelow= 4.7*2+answerinb)
Answered by
Tay
okay i got the answer to b.
now is a v(2)= vo+(9.8)(2)
or
v(2)= vo -(9.8)(2)?
now is a v(2)= vo+(9.8)(2)
or
v(2)= vo -(9.8)(2)?
Answered by
bobpursley
9.8 is a negative quantity, if up is +
Answered by
Tay
so the answer to a should be -14.9 m/s
and c should be -.08?
and c should be -.08?
Answered by
GRIFFIN
YES
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