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HELPMEPLEASE5 · Oct 4, 2009, 04:51 PM

A parachutist descending at a speed of 10.60 m/s loses a shoe at an altitude of 57.60 m.
(a) When does the shoe reach the ground?

twinkiedooter · Oct 4, 2009, 04:54 PM

Right before the body reaches the ground.

Perito · Oct 4, 2009, 05:45 PM

If you ignore friction caused by the air, bodies drop at the same time regardless of their mass. Therefore, the shoe will hit the ground about the same time the foot does. :rolleyes:

InfoJunkie4Life · Oct 4, 2009, 06:43 PM

Ok... they are trying to trick you on this one.

You are dropping a shoe from an elevation and it already has a starting speed of 10.6m/s.

These equation should help:
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$V_f = sqrt{V_i^2 + 2ax}$

$V_f$ is Final Velocity
$V_i$ is Initial Velocity
$a$ is Acceleration
$x$ is Distance
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$V_a = \frac{V_f + V_i}{2}$

$V_a$ is Average Velocity
$V_f$ is Final Velocity
$V_i$ is Initial Velocity
----------
$T = \frac{x}{V_a}$

$T$ is Time
$x$ is Distance
$V_a$ is Average Velocity
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Follow each equation with any information you have or that which you obtained from a previous equation.

That will tell you how long it took.

Alty · Oct 4, 2009, 06:46 PM

Just to remind everyone. We cannot do someone's homework for them.

OP, you have to show us the equation you came up with, how you solved it and your answer. We can then help you work through the problem.

Now that you have the equation, solve it, then come back for more help if you need it.

Unknown008 · Oct 5, 2009, 09:30 AM

There's another equation that will directly give you the time. That's:

$s = ut + \frac12 at^2$

s is the displacement, 57.6 m
u is the initial velocity, 10.6 m/s
a the acceleration due to gravity, 9.81 m/s^2

You'll end up with a quadratic equation in time 't'. Solve with the quadratic formula, and you're done. Note that time cannot be negative.

InfoJunkie4Life · Oct 5, 2009, 09:17 PM

I also think the following equation will work:

$t = \frac{sqrt{V_i^2 + 2ax} + V_i}{2}$

$V_i$ is Initial Velocity
$a$ is Acceleration
$x$ is Distance
$t$ is Time

InfoJunkie4Life · Oct 5, 2009, 09:23 PM

Sorry, last post was wrong.

its actually:

[math]t = \frac{2x}{\sqrt{V_i^2 + 2ax} + V_i}

InfoJunkie4Life · Oct 5, 2009, 09:24 PM

I'm getting worse at this...

Maybe I should use the preview option more. I think this is right now:

$t = \frac{2x}{\sqrt{V_i^2 + 2ax} + V_i}$

InfoJunkie4Life · Oct 5, 2009, 09:25 PM

I'm getting worse at this...

Maybe I should use the preview option more. I think this is right now:

$t = \frac{2x}{\sqrt{V_i^2 + 2ax} + V_i}$

Unknown008 · Oct 5, 2009, 11:24 PM

Oh my... for God's sake! Don't use that formula! This is so complicated, with fractions, square roots, power 2, etc!

The only ones you have to remember are:

$v = u+at$

$s = ut + \frac12 at^2$

$v^2= u^2+2as$

No square roots,no fraction of variable.

InfoJunkie4Life · Oct 5, 2009, 11:41 PM

I just took the available equations and put them together in a way that uses the available data so that you are able to just enter the known data and run a single calculation.

I know it isn't the greatest to do but it works.

I'd never expect any ever to remember it, I don't think I will ever after this topic.

Unknown008 · Oct 6, 2009, 12:03 AM

Lol! Then, please, don't put it like that. The simpler, the better. I'm reading one of your threads (a lengthy one) now, while I was looking for an old thread. I'll be answering soon.

InfoJunkie4Life · Oct 6, 2009, 12:09 AM

Ok... thanx