Answer to Question #229799 in Mechanics | Relativity for Gilbert

Question

A particle p is let fall vertically from a height to the ground. Another particle Qjs projected vertically down three seconds later with a velocity of 50 meter per seconds from the same height. If both particles reached the ground simultaneously,calculate the time each particle reach the ground and the distance traveled by each particle

Answer 1

The motion of the particles is the motion with a constant acceleration. Let us calculate the distance from the start of motion downwards. Let H be the initial height and t₁ be the time of motion, therefore $H = \dfrac{gt_1^2}{2}$ .

$t_1 = t_2 + 3\,\mathrm{s},$ and for the second particle $H = v_0t_2 + \dfrac{gt_2^2}{2}.$

$\dfrac{g(t_2+3)^2}{2} = v_0t_2 + \dfrac{gt_2^2}{2}, \\ 3gt_2 + 4.5 g = 50t_2, \\ t_2 = \dfrac{4.5g}{50-3g} \approx 2.15\,\mathrm{s}.$

The height H is $H = \dfrac{g(2.15+3)^2}{2} = 130\,\mathrm{m}.$