I made myself a question involving calculus and physics just for fun but can't figure it all out.

A rocket with mass $1000kg$ launches with a net force of $5000N$ for $5s$. Find the maximum height and the time taken to reach the peak.

First I found the velocity by integrating:

$$v = \int \frac{F}{m} \;\mathrm{dt}$$

Which yields:


and substituting everything in I find my maximum velocity: $$25ms^{-1}$$

Now once the velocity reached its maximum, the acceleration will become: $$-10ms^{-2}$$ and this is where my further calculations turn into magic. I figure it should be a bitwise function but can't piece it together.

I tried integrating the new acceleration and adding my maximum velocity as $c$, but using graphing tools I see it doesn't make sense. I reasoned that the graph should look like a parabola with the left "leg" being constant up to $25ms^{-1}$ followed by a normal "parabolic drop".

  • $\begingroup$ Now try a more realistic rocket, one that flies by ejecting mass. $\endgroup$ – PM 2Ring Oct 6 '18 at 14:51
  • $\begingroup$ Sounds fun, but don't I need differential equations or something like that to solve those? I haven't really tried one that eject mass as well $\endgroup$ – s8an-AB Oct 7 '18 at 21:52
  • $\begingroup$ Well, yes. But hey, it is rocket science. :) And the differential equation is pretty easy to integrate if you eject the reaction mass at a constant rate (constant exhaust velocity relative to the rocket). $\endgroup$ – PM 2Ring Oct 8 '18 at 4:33

Your solution needs two steps.

  1. Find height and velocity after the burn (time $t_1$)
    • The acceleration needs to account for thrust and gravity $a(t) = F(t)/m - g$ $$\begin{aligned} v(t) & = \int a(t)\,{\rm d}t=\int \limits_0^{t} \left( \frac{F}{m}-g \right) \,{\rm d}t = \left( \frac{F}{m}- g \right) t\\ & v_1 = \left( \frac{F}{m}- g \right) t_1 \end{aligned} $$ and $$ \begin{aligned} h(t) &= \int v(t)\,{\rm d}t =\int \limits_0^{t} \left( \frac{F}{m}-g \right) t\, {\rm d} t = \frac{t^2}{2}\left( \frac{F}{m}-g \right) \\ & h_1 = \frac{t_1^2}{2}\left( \frac{F}{m}-g \right) \end{aligned} $$
  2. Find maximum height during free fall (even though it is going upwards) $$ \begin{aligned} v(t) & = v_1 + \int \limits_{t_1}^t (-g)\, {\rm d}t = v_1 - g (t-t_1) \\ & v(t)=0 \left. \vphantom{\int } \right\} t_2 = t_1 + \frac{v_1}{g} \end{aligned} $$ and $$ \begin{aligned} h(t) &= h_1 + \int v(t)\,{\rm d}t = h_1 + \int \limits_{t_1}^t \left( v_1- g ( t-t_1) \right)\,{\rm d}t \\ & h_2 = h(t_2) = h_1 + \frac{v_1^2}{2 g} = \frac{t_1^2}{2 g} \frac{F}{m} \left( \frac{F}{m}-g \right) \end{aligned} $$
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    $\begingroup$ This looks more fun :) Will look at tommorow in more depth, brain needs its sleep at the moment. $\endgroup$ – s8an-AB Oct 3 '18 at 20:33

Gravitational force is approximately 9800N, so your rocket cannot even take off.

Edit: At the peak the velocity is zero, so you need to take the maximum velocity you already calculated $v_{max}$ and write down this equation:

$v_{max} + gt_p = 0$

where $t_p$ is the time it takes for the rocket to get from the point where it has maximum velocity to the peak height ($g$ is negative of course, it is acceleration provided by gravity). So the overall time to get to the peak is

$t_{overall} = t_p + t$

where $t$ is those initial 5 seconds. Can you continue from here?

  • $\begingroup$ Haha... I can't believe I didn't spot that. No net force in the upwards direction. I fixed the post. The net force upwards is 5000N $\endgroup$ – s8an-AB Oct 3 '18 at 19:37
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    $\begingroup$ 93.75m is the maximum height. $\endgroup$ – s8an-AB Oct 3 '18 at 20:04
  • $\begingroup$ That is correct :) $\endgroup$ – Andrej Oct 3 '18 at 20:08
  • $\begingroup$ Also found my bitwise function that describes the position too: $$s(t)=2.5t^2, t<=5$$ $$s(t)=-5t(t-5)^2+25(t-5)+62.5, t > 5$$ $\endgroup$ – s8an-AB Oct 3 '18 at 20:24

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