Girl falling out of helicopter - kinematics

[solars]

A girl in a helicopter falls out when it was rising vertically at 10 m/s. The girl has a rope tied to her harness so she can be retireved. After 2.5 secs, how much rope has been pulled out of the helicopter?

I know that this is essentially asking you to find the height that the helicopter traveled during the 2.5 sec, but i keep getting the wrong answer.

I used d=vt+0.5at^2
v=10
t=2.5

and i get a negative #! which is obviously wrong

 

[physicsnoob93]

Okay let me give u a hint, when the helicopter is rising up, the girl has the helicopter's velocity, at the instant she leaves the helicopter.

take note: Velocity also includes the direction.

In other words, she doesn't fall down, straightaway.

Get it? If you don't i'll explain further.

 

[ace123]

A negative number doesn't always mean your are wrong. Think about it. You made the up direction positive right? So the down direction is negative. Your displacement is from the helicopter to the bottom of the rope.

 

[solars]

A negative number doesn't always mean your are wrong. Think about it. You made the up direction positive right? So the down direction is negative. Your displacement is from the helicopter to the bottom of the rope.

Yah I see now, but I know the correct answer but wasn't able to get it, so that was why I knew it was wrong.

 

[solars]

Okay let me give u a hint, when the helicopter is rising up, the girl has the helicopter's velocity, at the instant she leaves the helicopter.

take note: Velocity also includes the direction.

In other words, she doesn't fall down, straightaway.

Get it? If you don't i'll explain further.

hmm i think i know where you are going with this. are you saying that the girl's initial velocity is zero then?

 

[physicsnoob93]

The girl's initial velocity is the same as the helicopters, not 0. Imagine it as a projectile, just that there is no movement in the "x direction". So she will keep going up until her velocity is 0, and then she will come down, due to the acceleration due to gravity.

 

[physicsnoob93]

There will be 2 parts to this, the first part where the Vi is the helicopter's velocity and since u took the up direction as positive, Vi is positive until it reaches a Vf which is 0, because at the top of it's path a projectile is at an instantaneous stop.
So you find the displacement for this first part, which would be the length of the rope released.

Next, from Vi = 0 to ground. where your acceleration is 9.81 (gravity). Calculate the displacement for this, and you should get the rope released, but I'm not sure because a releasing rope is a pretty weird qn and u don't know if the rope that the girl "went up with" will drag down or not. ><

But i hope this gives you a rough idea.

 

[alphysicist]

Hi solars,

A girl in a helicopter falls out when it was rising vertically at 10 m/s. The girl has a rope tied to her harness so she can be retireved. After 2.5 secs, how much rope has been pulled out of the helicopter?

I know that this is essentially asking you to find the height that the helicopter traveled during the 2.5 sec, but i keep getting the wrong answer.

I used d=vt+0.5at^2
v=10
t=2.5

and i get a negative #! which is obviously wrong

You've found out how far down the girl is from the point at which she fell out (after 2.5 seconds). So that's where one end of the rope is.

But the other end of the rope is at the helicopter. Where is the helicopter at 2.5 seconds after the girl has fallen? The distance between her positions and the helicopter's position is the length of rope released.