Homework Help: Grade 11 U Free fall question

1. Feb 19, 2009

Vaz17

Hey guys

need help beginning to solve the following:

A skydiver accidentally drops a camera out of the plane. The diver notices the mistake 3.0s later and dives out of the plane with a downward velocity of 10.0 m/s. The camera experiences free fall(9.8 m/s^2), but the sky diver accelerates downwards at 8.0m/s^2.

How far does the camera fall before the sky diver is able to catch it?

I found how far the camera falls in 3.0s= 44.1 m, and Vf of the camera is 29.4 m/s [down] since Vi is 0. I Don't know really where to go from here, any help will be appreciated!!

i think the teacher mentioned it takes 39 seconds, but how would i find that

2. Feb 19, 2009

djeitnstine

What is true about the displacement when the diver meets the camera?

3. Feb 19, 2009

Vaz17

displacement would be the same as the camera?

4. Feb 19, 2009

djeitnstine

precisely

5. Feb 19, 2009

Vaz17

so i need to do d=motion equation of camera=motion equation of diver?

6. Feb 19, 2009

djeitnstine

You've got it.

7. Feb 19, 2009

djeitnstine

If you've done it correctly you should get a quadratic

8. Feb 19, 2009

Vaz17

ok thank you

but how do i deal with the 10m/s and the 8.0 m/s^2 when deriving the divers motion equation? they can't simply be added together can they?

9. Feb 19, 2009

djeitnstine

I am assuming that you are using $$y=v_{i}t+ \frac{1}{2}at^{2}$$ where y is the vertical displacement

10. Feb 19, 2009

Vaz17

yes.

11. Feb 19, 2009

djeitnstine

Okay so setting 2 equations equal gives $$v_{i1}t+ \frac{1}{2}a_{1}t^{2}=v_{i2}t+\frac{1}{2}a_{2}t^{2}$$ which gives the quadratic $$(v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0$$

So no magic happens =)

Its up to you to decide which root is relevant

edit I left out my 1/2

12. Feb 19, 2009

Vaz17

thank you very much kind sir :)

13. Feb 19, 2009

djeitnstine

You are welcome

14. Feb 19, 2009

Vaz17

ok so what can i derive from that without t

edit where did you leave out the 1/2

15. Feb 19, 2009

djeitnstine

1) I placed the 1/2's that you see in the equation.

2) no t's should disappear since your velocities and accelerations are different

16. Feb 19, 2009

Fragment

A simple approach when comparing different velocities is using a table. Of course, I highly recommend working out the mathematics, but a table should also give you the same answer. Example: Car 1: 1s=30m, 2s=50m...Car 2: 1s=45m, 2s=49m....See where they interesect, which is like wording out a quadratic, so-to-speak.

Regards,

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17. Feb 19, 2009

Vaz17

hmmm

18. Feb 19, 2009

Fragment

The solution seems pretty clear to me from this point, djeit pretty much pointed it out. Either graph your quadratics, solve them algebraically, or use the table method.

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19. Feb 19, 2009

Vaz17

ok, I found the time, 4.5s, using the quadratic formula. it seems wrong, and once i have this time, how do i get displacement?

which variables do i plug into the displacement formula? my starting values?

20. Feb 19, 2009

djeitnstine

21. Feb 19, 2009

Vaz17

22. Feb 19, 2009

djeitnstine

no i did not add the 3 seconds...i was just showing the answer to the quadratic...remember $$\frac{-b+/- \sqrt{b^{2}-4ac}}{2a}$$

23. Feb 19, 2009

Vaz17

lol i think you misunderstood, i didnt know how to solve the equation you gave on pg 1

24. Feb 19, 2009

Fragment

Well, put in the numbers and let us see what you get.

25. Feb 19, 2009

ok:)