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Results tagged with Search options user 97

The time rate of change of the position of an object

3
votes
Your approach is correct, now simply add indices to everything, i.e. $$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$ and note that $t_2 = t_1 - 2\,\text{s}$. Then solve $15\,\t …
answered Oct 8 '14 by Tobias Kienzler
2
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.$$ Terminal velocity means that $a=0$. When you drop something, the forces acting on it are usually the constant$^\dagger$ gravitational force $\vec F_g\approx m\cdot\vec g$ and the velocity $v … $ dependent Drag. For "usual" objects in air and close to terminal velocity, it is the Newtonian drag $$\vec F_D = -\frac12\rho v^2C_DA\cdot \vec e_v,$$ pointing opposite to the velocity. Here $\rho$ is …
answered May 23 '13 by Tobias Kienzler