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I'm trying to solve this task. The body moves in a uniform field of gravity of the Earth. Resistance force the medium is proportional to the square of the velocity. In the initial moment of time, the body was at a height of H, and its speed was zero. To find the dependence of speed on time, speed on height, and height on time. I assume that I should use newton's second law and the conservation of energy law. But I'm stuck with formulas and don't understand. Is this the right way? I think the formula for velocity will be too complicated after solving

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    $\begingroup$ You are going right except for the last two steps, your differential equation is correct just integrate to find velocity as a function of time( the integral will include log) once you have velocity as function of time write velocity as dx/dt on integrating again you will get displacement as a function of time $\endgroup$
    – imposter
    Commented Jun 7, 2020 at 13:58
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    $\begingroup$ Hello there, and welcome to the Physics Stack Exchange! Homework and "check my work" questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. Please read this post on asking homework questions and this post for "check my work" questions. $\endgroup$
    – Vincent Thacker
    Commented Jun 7, 2020 at 14:02
  • $\begingroup$ energy is not conserved with friction/drag forces present. (Airline travel would be much cheaper if it were). $\endgroup$
    – JEB
    Commented Jun 7, 2020 at 14:51
  • $\begingroup$ The expression (1/2)at^2 for distance is valid only with constant acceleration. $\endgroup$
    – R.W. Bird
    Commented Jun 7, 2020 at 19:02

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You are on the right track to calculate $v(t)$ with the first two equations. Extract a factor of $k/m$ out of the rhs of the second equation, move some parts around and we have $$ \frac{dv}{gm/k-v^2} = \frac{k}{m}dt, $$ from which you can start the integration (take care of some minus sign btw) to obtain $v(t)$. After that, continue to integrate $v(t)$ w.r.t. $t$ to obtain $h(t)$. To have a quick check of your answer, see if $v(t)$ approaches the terminal velocity as $t \to \infty$, and see if $h(t)$ behaves as a linear function of $t$ as $t \to \infty$.

However, $v(h)$ is probably hard to obtain, as can be seen from the expression of $v(t)$ and $h(t)$ respectively.

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