scguy Posted May 28, 2005 Posted May 28, 2005 If a particle is fired up with a velocity ''u'' then ''t'' seconds later a second particle is fired up with the same initial velocity and from the same place, prove that the particles will meet after: (0.5t+ u/g) Where g is gravity. The equations used are just the basic equations of motion, i have gotten an answer similiar but i cant see how to prove this exactly.
rajama Posted May 28, 2005 Posted May 28, 2005 Their positions must coincide at time t2, so using x = vt + 0.5at^2: p1 will move to x = ut2-0.5gt2^2 & p2 will move to x = u(t2-t)-0.5g(t2-t)^2 = ut2 - ut - 0.5g(t2^2 + t^2 - 2tt2) equating these and canceling terms: 0 = u/g + 0.5t - t2 or t2 = 0.5t + u/g
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