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Old Oct 10th 2009, 04:50 AM   #21
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OOps !!!

I made a mistake with the decimal fraction, it should be 0.404694.

So 1 sec / ( 1-(sq-rt ( 1-0.404694)) = 4.3775 secs total time

= (1/2)at^2 = 0.5 x 9.8 x 4.3775^2 = 93.89 m

my mistake, your answer is correct.

Whenever this sort of question shows it's face again i'll memorize it using the following:

38m = 38 m/s + (1/2)g = v (final velocity)
= 38 + 4.9
v = gt
t= v/g = 42.9 m/s / 9.8 = 4.3775 secs

(1/2)gt^2 = 93.89 m

Last edited by Paul46; Oct 10th 2009 at 06:47 AM.
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Old Oct 10th 2009, 10:49 PM   #22
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I am kind of relieved the answer is correct, but i dont understand this bit

So 1 sec / (1-(sq-rt (1-0.35681)) = 5.05 secs total time falling

I know the 0.35 bit is wrong from your later post, but how do you come to this ratio at all ?

1 sec / (1-(sq-rt (1-0.35681))

Last edited by physicsquest; Oct 10th 2009 at 11:24 PM.
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Old Oct 11th 2009, 02:44 AM   #23
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physicsquest,

It's just a ratio to confirm the total time.

A stone is dropped from a cliff in the final second the stone travels 67m, find the the total displacement.

67m = v= 67 m/s + (1/2)g = 71.9 m/s
v=gt
t=v/g = 7.3367 secs
(1/2)gt^2 = 263.755m total displacement


To confirm is the answer was correct i would use:
67m = 0.254023 decimal fraction of total displacement.

final 1 sec/(1-sq-rt(1-0.254023)) = 7.3367 secs

But since now i know that the final velocity (v) will be displacement (67m) + (1/2g) i no longer need to confirm the answer using that formula. It may seem abit odd, but it's the way i like to drum it into myself.

Once i found your answer was correct i started breaking the calculations down to find a simple calculation of the answer, because i couldn't work the answer out, when i broke the calculations down i found the easy way to find (v) was displacement + (1/2)g then do the above calculations.

I hope this answers your question.
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Old Oct 11th 2009, 09:06 AM   #24
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Do I understand both of you right now, i.e. all three of us agree on the value of H? If so then I'm quite happy!
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Old Oct 11th 2009, 12:25 PM   #25
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Falling Object

Using the last second

S = 38 U is unknown, A is Gravity and T is one.

S = ut + 0.5At^2

U = 33.1 M / S at the beginning of the final second

V = U + AT U = 0, A is Gravity T is unknown to find the time up to point U = 33.1

T = 3.37 Seconds

T for the total journey is 4.37 seconds.

Substituting back into S = Ut + 0.5AT^2 with U = 0 A Gravity T = 4.37 gives

H = 93.89. Give or take a few millimeters depending on the flavour of G (9.8 used here)

G of 9.81 gives H as 93.84

This seems to confirm the number already calculated before.
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Old Oct 12th 2009, 12:09 PM   #26
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Just out of interest guys,

I've been thinking is there a way to solve the above sort of question using longer time durations. Tapping away on my calculator for a few hours i found how to do it.

A stone is dropped from a cliff, in the last 3 secs the stone travels 99m, find the total displacement.

u = initial velocity
v=final velocity
a=acceleration (9.8 m/s^2)
t=time
t1=given time (3)
s=displacement
s1= given displacement (99)

This is what i did to find it:

v = s1/t1 + 3a/2 = 33 + 14.7 = 47.7 m/s
t= v/a = 4.867 s (3 d.p)
s= ut + (1/2)at^2 = 116.07 (2 d.p)

Just to confirm:
4.867 s - 3 s = 1.867 s
(1/2)at^2 = 17.07
17.07 + 99 = 116.07

Last edited by Paul46; Oct 12th 2009 at 12:29 PM.
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