November 26th, 2016, 04:24 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  particles projected vertically
A stone is projected upwards vertically with initial speed u m/s; a second stone is thrown upward from the same point with the same initial speed t seconds later than the first one. Prove that they collide at a height of (4u^2g^2t^2)/(8g) meters above the point of projection. To work this out, I used uvast equations. For the first particle, u = u, s = h, a = g, and t = k; for the second particle, u = u, s = h, a = g and t = kt. The second particle has been in the air for t seconds less than the first particle; that's why its time = kt. k stands for the amount of time the first particle has been in the air. Anyway, I worked these out using s = ut + .5at^2 and I don't know where to go from here.
Last edited by skipjack; November 26th, 2016 at 09:38 AM. Reason: to include needed parentheses 
November 26th, 2016, 08:27 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,511 Thanks: 761 
answered at other site

November 26th, 2016, 10:24 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,063 Thanks: 1396 
If the stones collide at height h meters, h = uk  gk²/2 = u(k  t)  g(k  t)²/2, which simplifies, after dividing by t, to gt/2 = gk  u. Squaring gives g²t²/4 = g²k²  2guk + u², so 2guk  g²k² = u²  g²t²/4 = (4u²  g²t²)/4. Dividing by 2g gives uk  gk²/2 = (4u²  g²t²)/(8g), as desired. 

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