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

[Fragment]

Didn't you say you had all solved for (obviously except displacement)?
y=delta-d(displacement)

 

[Vaz17]

yes thank you, i just realized i can now sub in my time in each of the motion equations, the camera and diver.

 

[Vaz17]

y=delta-d(displacement)

can you explaint this formula please? i thought y=displacement? lol

 

[Fragment]

Sorry... I haven't bothered with LaTeX yet, basically what I typed was y=displacement, since the symbol for displacement is delta-d. Djeit's derivation had a mistake in it, try:

$$(v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0$$

 

[djeitnstine]

Hey I'm so sorry about my quadratic equation :S it was supposed to be $$(v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0$$

 

[Vaz17]

oh ok thank you guys very much

the original screwed me up so bad lol

 

[Fragment]

Since it's for your class, you should be able to do this yourself. Make sure you get a lot of practice, test time won't let you use PF

All we did was simply group the like terms together. Make sure you're comfortable with this, because a question like this is bound to be on one of your tests. (From experience)

Regards,

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[Vaz17]

haha you're right. And yes there may be one of these, for the most part we are doing slightly less difficult ones, without the 3 second start difference.

 

[Vaz17]

what answer did you get with the corrected quadratic?

i think you guys are missing the fact that the camera was falling for 3 seconds already and traveled 44.1 m...and neither of those numbers appeared in the quadratic...wow I am nowhere near close to finishing this lol

 

[Fragment]

I don't think I'm allowed to post an answer until you have first, that way I don't give out an easy solution. Did you manage to figure it out yourself?

Regards,

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[Vaz17]

i don't see why it would hurt, I would only gain from it.

but anyways, i got a time of 50 s from the quadratic, and need displacement still :(

im doing something wrong

 

[Fragment]

Can you post your work so we can see where you went wrong? 50s is not correct. It would be very helpful

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[Vaz17]

I don't know how to do those fancy equations so, this is the best i can do

(0-10)t + 1/2 (9.8-8)t^2=0
-10t + 0.9t^2=0

if you sub that into the quadratic you get 50 or another number if you use the other root...

i also want to know where the 3 second delay of the diver comes into play in these equations...because I am not seeing it

 

[Fragment]

$$(v_{i1}-v_{i2})t+\frac{1}{2}(a_{1}-a_{2})t^{2}=0$$

Your numbers are wrong, that's why the quadratic gives you 50...
$$v_1$$ here is the first velocity of the camera, and you had found that yourself in your first post (29.4), so I don't know how you got 0 there. As for $$a1$$ and $$a2$$, they are your accelerations of the camera and of the diver, respectively. (9.8 and 8.0). Try again with those numbers.



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[Vaz17]

wouldnt initial velocity of the camera be 0?

if v1 is 29.4, what's v2?

 

[Fragment]

10, since your diver had a velocity of 10.0m/s when he dove out of the plane...

 

[Vaz17]

ok i see,

v1 should still be 0, as 29.4 was its the cameras final velocity.

I got 22 s

 

[Fragment]

That's not your final velocity. You are comparing velocities and accelerations of two objects in free-fall in an attempt to see where they cross. So 29.4m/s is the velocity that your camera had (at 3.0s) when the diver dove from the place. Thus it is clearly $$v_1$$. Otherwise, let it be 0, and you will notice that $$v_2$$ also needs to be 0. This is where the 3.0s comes into play. I hope you see the logic in this.



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[Vaz17]

I think I get it, but the diver isn't in free fall...

was 22 s correct?

 

[Fragment]

Correct.

 

[Vaz17]

I don't need to add the 3s?

and to get my displacement, do i use the numbers from the simplified quadratic in the displacement formula?

 

[Fragment]

For the 3s, no. You don't need to because as I said, you're comparing the times. Our 22s says it takes 22s of free-fall for the objects to cross paths. The 3s was already accounted for when we supposed that $$v_1$$ was 29.4m/s.

For the displacement, you managed to figure out time, and since you know that,
$$y=v_{i}t+ \frac{1}{2}at^{2}$$

You have your answer really, 22s tells you everything. Go ahead and try putting 22s in both $$y_1$$ and $$y_2$$, see what you get for both displacements...

 

[Fragment]

Do you clearly understand everything we have done so far? It was pretty simple; in physics you need to pay good attention to what is happening, and to what numbers to use and where.

 

[Vaz17]

ok but like i said, not both objects are in free fall, only the camera..right?

 

[Fragment]

What is your diver doing then?

 

[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?