The mass of rocket without fuel is m, and the mass of fuel is m.
Since it is high school physics, we should use high school mathematics to analyze it, but let's make a hypothesis first, that is, the impulse generated by mega launching fuel (I don't know if I remember it correctly, it should be called this, I=F*t=mv, and the impulse is equal to the force multiplied by the time of the force, which is equal to the momentum change of the object). Compared with the impulse generated by gravity resistance, the impulse generated by gravity and resistance can be ignored, that is, during the acceleration of the rocket.
Let's make another assumption, that is, every time the rocket ejects m/n weight of fuel, we assume that n is a very large number.
In this way, the following conclusions can be summarized.
The number of times-the momentum of the ejected fuel-is converted into the momentum of the rocket and the remaining fuel.
1 mv/n = [m+(n-1) m/n] * v 1 kloc-0/the fuel injected at the first time increased the rocket speed by v1.
The fuel relative to the ground speed is v.
The second time m (v-v1)/n = [m+(n-2) m/n] * V2 The fuel ejected for the second time increased the rocket speed by v2.
The fuel relative to the ground speed is v-V 1.
The third time m (v-v1-V2)/n = [m+(n-2) m/n] * V2 The fuel injected for the second time increased V2' s rocket speed.
The fuel relative to the ground speed is v-V 1-V2.
etc
The nth m (v-v 1-v2...-v
I won't say much about solving this mathematical formula. Everyone in high school should be able to do it. If you want to be strict, you can also prove it by mathematical induction (hehe)
The end result is
The acceleration speed of the rocket is v0 = v1+v2+...+VN = mv/[m+(n-1) m/n].
If n tends to infinity, then the rocket speed is V0=mv/(M+m).
Then use this speed to calculate the height of the rocket, and I won't count it.
The actual speed is like this for two reasons. First, the energy of fuel is not only transferred to the rocket, but also to the fuel that has not been ejected. Second, the fuel injection speed-the ground speed is reduced.
Now, to put it more realistically, it takes a relatively long time to fuel in mega launch. At this time, the influence of gravity and resistance on the acceleration process of rocket can not be ignored. In this case, the maximum speed of the rocket is even smaller than the above. Considering the influence of gravity and resistance, it is difficult to solve it with elementary mathematics, and calculus can easily get the answer. I'll mention it here.
Considering that fuel injection takes a certain time, it is assumed that fuel injection with m mass takes t time, and the fuel injection rate k = m/t.
Suppose the rocket accelerates to V(t) in the time from T to t+dt, and the fuel quality keeps m(t)=m-k*t(dt stands for a small time interval).
The analysis can be summarized as follows:
When the fuel ejects k*dt, its speed to the ground is v-V(t), which will increase the speed of the rocket by dV(t).
Considering the influence of gravity, the following equation can be obtained.
k * dt *[V-V(t)]=[M+M(t)]* dV(t)+[M+M(t)]( 1+p %)dt
The left side of the equation represents the momentum generated by injecting fuel with a weight of k*dt. The first term on the right side of the equation represents the value that the momentum of the fuel increases the momentum of the rocket, and the second term is the impulse that the momentum of the fuel is used to offset gravity and resistance.
Solve this differential equation, and add the initial velocity of zero and the value of acceleration [kv/(m+M)-( 1+p%)g] to get the final result. Because the derivation process is successful, I won't write it. If it is wrong, please help me correct it, hehe.
Speed at time t (acceleration period)
v(t)= k * v * t/(M+M)-( 1+p %)gt-(M+M-k * t){ ln[M+M-k * t)/(M+M)]- 1 }( 1+p %)g/k
Substitute t=T=m/k to get the speed when the rocket accelerates.
VM = mv/(M+M)-{ M *[ln(M/(M+M)- 1]+M } *( 1+p %)g/k
The first term on the right side of the equation is the result of not considering gravity resistance, and the latter term is the result of considering gravity and resistance.
I won't calculate the height either, but it should be noted that the height of the rocket in this case should be the sum of the height when accelerating and the height when going up freely. The speed of the rocket acceleration process has been calculated before, and the height can be known by integrating it.
There is an interesting conclusion. You can see the expression of V(t) above. If the first item, that is, the speed of fuel injection is not high enough, the rocket will never fly into the sky (hehe).
It should also be pointed out that in actual operation, the resistance is related to speed, but not to weight. Sometimes it can be assumed that the resistance f=K*V (the resistance speed is proportional to the first power), so the problem is more complicated. If anyone is interested, you can calculate it:)