<![CDATA[Math Is Fun Forum / Time and velocity]]> 2016-09-16T09:12:30Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=23381 <![CDATA[Re: Time and velocity]]>

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http://www.mathisfunforum.com/profile.php?id=212486 2016-09-16T09:12:30Z http://www.mathisfunforum.com/viewtopic.php?pid=387855#p387855
<![CDATA[Re: Time and velocity]]> hi peter010

question wrote:

It's height is h after t1 and t2 seconds

Just one h for both times.  Once on the way up and then again on the way down.

Bob

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http://www.mathisfunforum.com/profile.php?id=67694 2016-09-13T14:57:14Z http://www.mathisfunforum.com/viewtopic.php?pid=387768#p387768
<![CDATA[Re: Time and velocity]]> bob bundy wrote:

hi markosheehan

Use

a = -g, Write the two equations for h with t1 and t2.  Equate to get an expression for u in terms of g, t1 and t2.  Substitute back into either to eliminate u.

Bob

ps.  Why did I do it this way?  The problem involves height, gravity, initial velocity and time so that forces which equation to use.  With two equations you can eliminate one variable.  I chose h because it looked easiest but then had to rework to eliminate u.  Maybe I should have made u the subject of the two and equated those to get the answer straight off.  I leave it to the reader to try that if desired. pps.  Tried it myself and it comes out easily this way in four lines.

Hi,

I followed that, but we still have then h1 (at t1) and h2 (at t2)..

[t1.h2-t2.h1]/t1.t2=0.5g[t1-t2]

what do you think ?

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http://www.mathisfunforum.com/profile.php?id=212962 2016-09-13T08:51:48Z http://www.mathisfunforum.com/viewtopic.php?pid=387763#p387763
<![CDATA[Re: Time and velocity]]> I got this/
final t is comprised of  t1 of t2 times .. in other words -> t_final=t1*t2  (considering that t2 is just a unit less number)
-- Applying motion equation:
v= u-gt // minus because its going up
but v when particle reaches the top = 0
0=u-gt
u=gt..
lets integrate this over time..
h=0.5gt^2
//substitute: t=t1*t2
h=0.5g[t1*t2]^2
//arranging the equation..
[t1*t2]^2 = 2h/g   // but i know its not correct since units are not compatible on both sides-> unless considering the whole set of t1*t2 is of second's unit but not second square, as pre my assumption t=t1*t2 !
then i can say:
t1*t2 = 2h/g

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http://www.mathisfunforum.com/profile.php?id=212962 2016-09-13T08:29:21Z http://www.mathisfunforum.com/viewtopic.php?pid=387762#p387762
<![CDATA[Re: Time and velocity]]> hi markosheehan

Use

a = -g, Write the two equations for h with t1 and t2.  Equate to get an expression for u in terms of g, t1 and t2.  Substitute back into either to eliminate u.

Bob

ps.  Why did I do it this way?  The problem involves height, gravity, initial velocity and time so that forces which equation to use.  With two equations you can eliminate one variable.  I chose h because it looked easiest but then had to rework to eliminate u.  Maybe I should have made u the subject of the two and equated those to get the answer straight off.  I leave it to the reader to try that if desired. pps.  Tried it myself and it comes out easily this way in four lines.

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http://www.mathisfunforum.com/profile.php?id=67694 2016-09-13T07:43:52Z http://www.mathisfunforum.com/viewtopic.php?pid=387759#p387759
<![CDATA[Time and velocity]]> A particle is projected vertically upwards with velocity u m/s . It's height is h after t1 and t2 seconds. Prove that.       t1× t2 = 2h/g

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http://www.mathisfunforum.com/profile.php?id=212664 2016-09-12T19:59:35Z http://www.mathisfunforum.com/viewtopic.php?pid=387745#p387745