Math Is Fun Forum

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

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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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

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

peter010

Member

Registered: 2016-09-05

Posts: 22

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.

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 ?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Time and velocity

hi peter010

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

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Time and velocity

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{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.}