Grade 11 U Free fall question

In summary, the problem is about a skydiver who 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 8.0 m/s^2. The question is how far does the camera fall before the sky diver is able to catch it
  • #1
Vaz17
39
0
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
 
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  • #2
What is true about the displacement when the diver meets the camera?
 
  • #3
displacement would be the same as the camera?
 
  • #4
precisely
 
  • #5
so i need to do d=motion equation of camera=motion equation of diver?
 
  • #6
You've got it.
 
  • #7
If you've done it correctly you should get a quadratic
 
  • #8
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
I am assuming that you are using [tex]y=v_{i}t+ \frac{1}{2}at^{2}[/tex] where y is the vertical displacement
 
  • #10
yes.
 
  • #11
Okay so setting 2 equations equal gives [tex]v_{i1}t+ \frac{1}{2}a_{1}t^{2}=v_{i2}t+\frac{1}{2}a_{2}t^{2}[/tex] which gives the quadratic [tex](v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0[/tex]

So no magic happens =)

Its up to you to decide which root is relevant

edit I left out my 1/2
 
  • #12
thank you very much kind sir :)
 
  • #13
You are welcome
 
  • #14
ok so what can i derive from that without t

edit where did you leave out the 1/2
 
  • #15
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
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,

Fragment
 
  • #17
hmmm
 
  • #18
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.:smile:

Fragment
 
  • #19
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
rework your quadratic...i got 10.7777s
 
  • #22
no i did not add the 3 seconds...i was just showing the answer to the quadratic...remember [tex]\frac{-b+/- \sqrt{b^{2}-4ac}}{2a}[/tex]
 
  • #23
lol i think you misunderstood, i didnt know how to solve the equation you gave on pg 1
 
  • #24
Well, put in the numbers and let us see what you get.:smile:
 
  • #25
ok:)
 
  • #26
LaTeX Code: (v_{i1}-v_{i2})t+(a_{1}-a_{2}t^{2})t^{2}=0

when i plug in the values for this, i get 37.2...:S
 
  • #27
I don't know what to do after that
 
  • #28
In that case, you should create a post in the math forum asking how to solve a quadratic equation. Also you did not read my equation carefully as the result is not a single number.
 
  • #29
You should fix your LaTeX code though, it would help me read what you plugged in. After that we might be able to see where you struggle.

Regards,

Fragment
 
  • #30
no i know how to solve a quadratic, I am just confused with your equation, i think that's the basis of my whole incorrect solution right now lol.

fragment, that code was for djeitnstine's equation on pg 1.
 
  • #31
ok i got it!

now how do i solve for displacement?
 
  • #32
Remember that y is your displacement. What did you do by solving your quadratic?
 
  • #33
i solved for time, a little differently then your formula, check the link in one of my above posts.
 
  • #34
And so, where are you now in your problem?
 
  • #35
I need displacement given the time it took for the diver to reach the camera.
 

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