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Old Mar 21st 2018, 04:20 AM   #1
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Join Date: Mar 2018
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1D Kinematics

Motion
An object falls a distance h from rest.
If it travels half the height h in the last 1.00 s, find:
(a) the time
(b) the height h of its fall.
(c) Explain the physically unacceptable solution of the
quadratic equation in t that you obtain.

I am trying to use the:

{y=y(initial) + v(initial) - 1/2gt^2}

equation to for example say that y = h/2 and y(initial) = h, assuming that the v(initial) is also 0m/s and g=9.81. Also, I am trying to apply the equation to the case where y = 0 and y(initial) = h/2, where t = 1s and v(initial) is unknown.

I just can't seem to get the right answer. Thanks all.
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Old Mar 21st 2018, 05:25 AM   #2
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S = Ut +1/2f t^2 ................. S is distance traveled ........ U is initial velocity

Let T be the total time of fall ....

time for first part of fall is T-1 .... distance for first part of fall is h/2 ...initial velocity for first part of fall is 0

so ..... h/2= 0 + g/2 * (T-1)^2

............................xxxxxxxxxxxxxxxxxxxx.. ...................

whole fall s=h ....time of fall =T....initial velocity =0

so ..... h =0+ g/2 T^2

combining these two .... g/2 T^2 = 2*[ g/2 (T-1)^2]

solving that should get the answers
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