How Far Does the Camera Fall Before the Skydiver Catches It?

[Vaz17]

8.0 m/s^2

free fall is 9.8

or does the 8.0 reach 9.8 anyway

what do you mean y2...in that equation there is only y

not sure what you meant by that

 

[Fragment]

Ok, free-fall is any object not supported by anything, it is falling in the air, even if it was accelerating at 0.000001 m/s^2. They are BOTH in free-fall, just at different velocities. The diver will reach the camera because he had an initial velocity of 10.0m/s, and was accelerating by 8.0 m/s every second, so (8+10), then (8+8+10) and so on...
I strongly advise you look over your notes.

$$v_1$$ implies the equation for the camera, and 2 implies for the diver...
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[Vaz17]

lol

i understand now, never had that def of free fall in my notes surprisingly.

thank you

 

[Fragment]

9.8 is the gravitational constant as well as the acceleration due to gravity on Earth (they're the same thing anyway). So since they are both in free-fall, we used both equations together to solve for time(substitution). After we had found time we could use it for find displacement, which I hope you see how to do...(Hint: Use the very first equations, see page.1)

Regards,

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[Vaz17]

644.6 m

 

[Fragment]

Can you show your work please? I want to be sure you're using the right numbers.
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[Vaz17]

i subd in 22, 19.4, and .9 (acceleration)

 

[Fragment]

In , $$y=v_{i}t+ \frac{1}{2}at^{2}$$?

 

[Vaz17]

yes.

 

[Fragment]

You aren't thinking of what is happening. 19.4 and 0.9 are not the numbers you need to use right now. Think about: What is the velocity of the camera, what is its acceleration?

 

[Vaz17]

oh, we need to see what is the displacement given the time, as it is the same for both!

 

[Fragment]

Exactly, this was your ultimate goal from the start. Where they intersect will be where they have the same displacement at a given time, which we had to solve for using the quadratic formula. So:

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.

Using:
$$<br /> y=v_{i}t+ \frac{1}{2}at^{2}<br />$$

Tell me what you get.

 

[Vaz17]

3018.4 m

 

[Fragment]

Good. Now that you have an answer, do you understand the steps we've went through? Can you duplicate this for another question? Having the right answer is good, but pointless if you don't understand how you got it. Revise your work and understand it, you'll come off top of your class if you're able to do this.
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[Vaz17]

believe me, after your step by step help, and me watching my mistakes and learning from them, i know what to look for in future similar questions.

thank you

 

[Fragment]

You're very welcome, if you have any more problems that you need help, or just to double check, post them in this thread and I'll gladly help you along.

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[Vaz17]

will do, thanks!