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November 26th, 2016, 05:24 AM   #1
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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 10:38 AM. Reason: to include needed parentheses
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November 26th, 2016, 09:27 AM   #2
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November 26th, 2016, 11:24 AM   #3
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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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