Math Is Fun Forum

markosheehan

Member

Registered: 2016-06-15

Posts: 51

projectile

a particle is projected vertically upwards from a point P. at the same instant a second particle is let fall  from rest vertically at q. q is directly above p. the 2 particles collide at a point r after t seconds . when the 2 particles collide they are travelling at equal speeds. prove that |pr|=3|rq|

i am trying to solve this with uvast equations   for the first particle i have v=v   s=r   a=-g  t=t    for the second particle i have  v=v   s=p-r   a=g  t=t dont know where to go from here

Last edited by markosheehan (2016-12-18 05:22:53)

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: projectile

You are on correct path. eliminate t from the 2 equations.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
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zetafunc

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Registered: 2014-05-21

Posts: 2,436

Website

Re: projectile

250y=250(tan a)x-(1-tan^2 a)x^2

Looks to me like it should be

instead -- apart from that, as thickhead says, just kill the

using your expression for

and it works out nicely.

Last edited by zetafunc (2016-12-18 05:33:10)

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: projectile

For the new problem get u in terms of gt and calculate s1 and s2. Simple. I am afraid I should not reveal anything more.

---

{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.}

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

Website

Re: projectile

In future, I would recommend creating a new thread for your new problem, instead of editing your original post, deleting the old question and replacing it with a new one. That way, other people can benefit, too.

Last edited by zetafunc (2016-12-19 00:53:29)