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## #1 2016-09-12 07:59:35

markosheehan
Member
Registered: 2016-06-15
Posts: 51

### 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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## #2 2016-09-12 19:43:52

Bob
Registered: 2010-06-20
Posts: 9,021

### 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.

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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## #3 2016-09-12 20:29:21

peter010
Member
Registered: 2016-09-05
Posts: 22

### 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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## #4 2016-09-12 20:51:48

peter010
Member
Registered: 2016-09-05
Posts: 22

### 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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## #5 2016-09-13 02:57:14

Bob
Registered: 2010-06-20
Posts: 9,021

### 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

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

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## #6 2016-09-15 21:12:30

Member
Registered: 2016-04-16
Posts: 1,086

### Re: Time and velocity

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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