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 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^2-g^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 = k-t. The second particle has been in the air for t seconds less than the first particle; that's why its time = k-t. 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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