# Grade 11 U Free fall question

• Vaz17
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
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

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

displacement would be the same as the camera?

precisely

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

You've got it.

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

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?

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

yes.

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

thank you very much kind sir :)

You are welcome

ok so what can i derive from that without t

edit where did you leave out the 1/2

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

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

hmmm

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.

Fragment

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?

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

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

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

ok:)

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

I don't know what to do after that

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.

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

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.

ok i got it!

now how do i solve for displacement?

i solved for time, a little differently then your formula, check the link in one of my above posts.

And so, where are you now in your problem?

I need displacement given the time it took for the diver to reach the camera.

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