1. ## Motion problems

Can someone please show me the steps needed to achieve the answers to this question?

A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

I know that's going to be a lot of work, but any help is much appreciated. =)

2. Originally Posted by Hellooo
Can someone please show me the steps needed to achieve the answers to this question?

A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

I know that's going to be a lot of work, but any help is much appreciated. =)
What have you tried? Where are you stuck?

3. I'm completely stuck on the question.

4. Originally Posted by Hellooo
Can someone please show me the steps needed to achieve the answers to this question?

A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

I know that's going to be a lot of work, but any help is much appreciated. =)
Package: Initial velocity = 30 m/s, acceleration = -9.8 m/s^2, t = 2 s, Displacement = x1 = ?

Helicopter: Iinitial velcocity = 30 m/s, acceleration = 5.2 m/s^2, t = 2 s, Displacement = x2 = ?

Distance between parcel and helicopter = x2 + 5 - x1 (the 5 comes from the rope)

5. So dx=30m, correct? Thank you very much for the quick replies =) and sorry for double posting.

6. Originally Posted by Hellooo
So dx=30m, correct? [snip]

7. dy1=0.5(-9.8)(2)^2+30(2)
=40.4m

dy2=0.5(5.2)(2)^2+40(2)
=91.2m

dy2=91.2+5=96.2m

dy=96.2m-40.4m=50.8m?

8. Originally Posted by Hellooo
dy1=0.5(-9.8)(2)^2+30(2)
=40.4m Mr F says: Correct.

dy2=0.5(5.2)(2)^2+40(2)
=91.2m Mr F says: No. What is the 40 doing there? Fix this, put it into the working below and you will get the correct answer.

dy2=91.2+5=96.2m

dy=96.2m-40.4m=50.8m?
..

9. i thought we use v2 to find the displacement of the helicopter, v2=5.2 x2 +30=40.4m/s up. Is that incorrect? Thanks for being patient with me =)

10. Originally Posted by Hellooo
i thought we use v2 to find the displacement of the helicopter, v2=5.2 x2 +30=40.4m/s up. Is that incorrect? Thanks for being patient with me =)
You're using a fomula that looks something like $\displaystyle x = v_0 t + \frac{1}{2} a t^2$. Do you understand the meaning of each symbol in this formula? In particular, do you know what $\displaystyle v_0$ means?

11. oh sorry for my misunderstanding, so dy2 should be 0.5(5.2)(2)^2+30(2)=70.4? Thanks for correcting me once again

12. Originally Posted by Hellooo
oh sorry for my misunderstanding, so dy2 should be 0.5(5.2)(2)^2+30(2)=70.4? Thanks for correcting me once again
Yes.